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Approximately How Long Does It Take Until the Soccer Ball Hits the Ground Again?

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Learning Objectives

By the terminate of this section, you will be able to:

  • Apply one-dimensional motion in perpendicular directions to clarify projectile motion.
  • Summate the range, time of flying, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
  • Detect the time of flying and touch velocity of a projectile that lands at a different top from that of launch.
  • Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, bailiwick but to acceleration every bit a issue of gravity. The applications of projectile movement in physics and engineering are numerous. Some examples include meteors as they enter Earth'due south atmosphere, fireworks, and the motion of whatever brawl in sports. Such objects are called projectiles and their path is chosen a trajectory. The motion of falling objects as discussed in Motion Along a Straight Line is a simple i-dimensional blazon of projectile motion in which there is no horizontal motility. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The well-nigh important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: 1 forth the horizontal axis and the other along the vertical. (This choice of axes is the most sensible considering acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, nosotros call the horizontal centrality the ten-axis and the vertical axis the y-centrality. Information technology is not required that we utilize this choice of axes; it is simply convenient in the instance of gravitational acceleration. In other cases we may choose a different set of axes. (Figure) illustrates the note for displacement, where we define [latex] \overset{\to }{south} [/latex] to be the total displacement, and [latex] \overset{\to }{10} [/latex] and [latex] \overset{\to }{y} [/latex] are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are southward, x, and y.

An illustration of a soccer player kicking a ball. The soccer player's foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle theta between the x axis and s.

Figure 4.xi The total deportation s of a soccer brawl at a betoken along its path. The vector [latex] \overset{\to }{s} [/latex] has components [latex] \overset{\to }{10} [/latex] and [latex] \overset{\to }{y} [/latex] along the horizontal and vertical axes. Its magnitude is s and information technology makes an angle θ with the horizontal.

To describe projectile motion completely, nosotros must include velocity and dispatch, too as displacement. We must notice their components along the ten- and y-axes. Let's assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

[latex] {a}_{y}=\text{−}one thousand=-9.8\,\text{m}\text{/}{\text{southward}}^{ii}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}). [/latex]

Because gravity is vertical, [latex] {a}_{x}=0. [/latex] If [latex] {a}_{x}=0, [/latex] this means the initial velocity in the x direction is equal to the terminal velocity in the 10 management, or [latex] {5}_{x}={five}_{0x}. [/latex] With these weather on acceleration and velocity, we can write the kinematic (Equation) through (Equation) for motion in a uniform gravitational field, including the residual of the kinematic equations for a abiding acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with [latex] {a}_{y}=\text{−}thou,\enspace{a}_{x}=0: [/latex]

Horizontal Motion

[latex] {v}_{0x}={v}_{x},\,x={10}_{0}+{five}_{10}t [/latex]

Vertical Motion

[latex] y={y}_{0}+\frac{ane}{two}({v}_{0y}+{5}_{y})t [/latex]

[latex] {5}_{y}={v}_{0y}-gt [/latex]

[latex] y={y}_{0}+{v}_{0y}t-\frac{1}{2}g{t}^{two} [/latex]

[latex] {5}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0}) [/latex]

Using this set of equations, nosotros tin analyze projectile move, keeping in mind some important points.

Problem-Solving Strategy: Projectile Movement

  1. Resolve the motility into horizontal and vertical components along the 10– and y-axes. The magnitudes of the components of displacement [latex] \overset{\to }{s} [/latex] along these axes are x and y. The magnitudes of the components of velocity [latex] \overset{\to }{five} [/latex] are [latex] {v}_{x}=v\text{cos}\,\theta \,\text{and}\,{v}_{y}=five\text{sin}\,\theta , [/latex] where five is the magnitude of the velocity and θ is its direction relative to the horizontal, equally shown in (Figure).
  2. Care for the motion equally two contained ane-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical movement presented earlier.
  3. Solve for the unknowns in the 2 separate motions: 1 horizontal and i vertical. Note that the merely common variable between the motions is fourth dimension t. The trouble-solving procedures here are the same equally those for 1-dimensional kinematics and are illustrated in the following solved examples.
  4. Recombine quantities in the horizontal and vertical directions to find the full displacement [latex] \overset{\to }{south} [/latex] and velocity [latex] \overset{\to }{five}. [/latex] Solve for the magnitude and direction of the deportation and velocity using

    [latex] s=\sqrt{{x}^{ii}+{y}^{2}},\enspace\theta ={\text{tan}}^{-i}(y\text{/}x),\enspace{v}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}, [/latex]

    where θ is the direction of the displacement [latex] \overset{\to }{southward}. [/latex]

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile's position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component.

Figure 4.12 (a) We analyze ii-dimensional projectile motion by breaking information technology into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because [latex] {a}_{x}=0 [/latex] and [latex] {v}_{x} [/latex] is a abiding. (c) The velocity in the vertical management begins to decrease as the object rises. At its highest point, the vertical velocity is goose egg. Equally the object falls toward Earth again, the vertical velocity increases again in magnitude but points in the reverse direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given betoken on the trajectory.

Example

A Fireworks Projectile Explodes Loftier and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 thou/s at an angle of [latex] 75.0\text{°} [/latex] above the horizontal, as illustrated in (Figure). The fuse is timed to ignite the shell just equally it reaches its highest signal higher up the ground. (a) Calculate the height at which the crush explodes. (b) How much time passes betwixt the launch of the shell and the explosion? (c) What is the horizontal displacement of the crush when it explodes? (d) What is the total displacement from the signal of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees.

Figure 4.xiii The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest signal in its trajectory, which is constitute to be at a elevation of 233 chiliad and 125 grand abroad horizontally.

Strategy

The motility tin exist broken into horizontal and vertical motions in which [latex] {a}_{x}=0 [/latex] and [latex] {a}_{y}=\text{−}g. [/latex] We tin and then ascertain [latex] {ten}_{0} [/latex] and [latex] {y}_{0} [/latex] to be zero and solve for the desired quantities.

Solution

(a) By "height" we hateful the altitude or vertical position y higher up the starting indicate. The highest point in whatever trajectory, called the apex, is reached when [latex] {v}_{y}=0. [/latex] Since we know the initial and final velocities, besides as the initial position, nosotros apply the following equation to notice y:

[latex] {v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0}). [/latex]

Because [latex] {y}_{0} [/latex] and [latex] {v}_{y} [/latex] are both zero, the equation simplifies to

[latex] \text{0}={v}_{0y}^{ii}-2gy. [/latex]

Solving for y gives

[latex] y=\frac{{5}_{0y}^{ii}}{2g}. [/latex]

Now we must find [latex] {v}_{0y}, [/latex] the component of the initial velocity in the y management. It is given by [latex] {v}_{0y}={v}_{0}\text{sin}{\theta }_{0}, [/latex] where [latex] {v}_{0} [/latex] is the initial velocity of seventy.0 m/s and [latex] {\theta }_{0}=75\text{°} [/latex] is the initial angle. Thus,

[latex] {five}_{0y}={5}_{0}\text{sin}\,\theta =(70.0\,\text{grand}\text{/}\text{due south})\text{sin}\,75\text{°}=67.6\,\text{chiliad}\text{/}\text{south} [/latex]

and y is

[latex] y=\frac{{(67.half-dozen\,\text{m}\text{/}\text{s})}^{2}}{two(9.80\,\text{m}\text{/}{\text{s}}^{2})}. [/latex]

Thus, we take

[latex] y=233\,\text{m}\text{.} [/latex]

Notation that considering upwards is positive, the initial vertical velocity is positive, as is the maximum pinnacle, merely the dispatch resulting from gravity is negative. Note also that the maximum pinnacle depends just on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum pinnacle of 233 g (neglecting air resistance). The numbers in this case are reasonable for big fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is non completely negligible, then the initial velocity would accept to be somewhat larger than that given to reach the same height.

(b) As in many physics bug, there is more than than one mode to solve for the time the projectile reaches its highest indicate. In this instance, the easiest method is to use [latex] {v}_{y}={v}_{0y}-gt. [/latex] Because [latex] {five}_{y}=0 [/latex] at the apex, this equation reduces to simply

[latex] 0={v}_{0y}-gt [/latex]

or

[latex] t=\frac{{v}_{0y}}{g}=\frac{67.vi\,\text{grand}\text{/}\text{southward}}{9.80\,\text{m}\text{/}{\text{southward}}^{2}}=6.xc\text{s}\text{.} [/latex]

This time is also reasonable for large fireworks. If yous are able to come across the launch of fireworks, observe that several seconds pass before the vanquish explodes. Some other style of finding the time is by using[latex] y\,\text{=}\,{y}_{0}+\frac{one}{two}({v}_{0y}+{five}_{y})t. [/latex] This is left for you as an exercise to complete.

(c) Because air resistance is negligible, [latex] {a}_{x}=0 [/latex] and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time every bit given by [latex] x={x}_{0}+{v}_{x}t, [/latex] where [latex] {ten}_{0} [/latex] is equal to zero. Thus,

[latex] ten={v}_{10}t, [/latex]

where [latex] {v}_{x} [/latex] is the x-component of the velocity, which is given by

[latex] {v}_{x}={v}_{0}\text{cos}\,\theta =(70.0\,\text{thou}\text{/}\text{s})\text{cos}75\text{°}=eighteen.1\,\text{chiliad}\text{/}\text{s}. [/latex]

Time t for both motions is the aforementioned, and so x is

[latex] 10=(18.ane\,\text{one thousand}\text{/}\text{southward})6.90\,\text{s}=125\,\text{m}\text{.} [/latex]

Horizontal movement is a constant velocity in the absence of air resistance. The horizontal deportation institute hither could be useful in keeping the fireworks fragments from falling on spectators. When the trounce explodes, air resistance has a major effect, and many fragments country directly below.

(d) The horizontal and vertical components of the displacement were merely calculated, so all that is needed here is to detect the magnitude and direction of the deportation at the highest point:

[latex] \overset{\to }{s}=125\chapeau{i}+233\hat{j} [/latex]

[latex] |\overset{\to }{s}|=\sqrt{{125}^{2}+{233}^{two}}=264\,\text{m} [/latex]

[latex] \theta ={\text{tan}}^{-i}(\frac{233}{125})=61.8\text{°}. [/latex]

Notation that the angle for the deportation vector is less than the initial bending of launch. To see why this is, review (Figure), which shows the curvature of the trajectory toward the ground level.

When solving (Effigy)(a), the expression nosotros found for y is valid for whatever projectile motion when air resistance is negligible. Call the maximum height y = h. Then,

[latex] h=\frac{{5}_{0y}^{ii}}{2g}. [/latex]

This equation defines the maximum height of a projectile in a higher place its launch position and it depends merely on the vertical component of the initial velocity.

Check Your Understanding

A rock is thrown horizontally off a cliff [latex] 100.0\,\text{m} [/latex] high with a velocity of 15.0 m/southward. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations draw the vertical motility? (d) What is the rock's velocity at the bespeak of impact?

(a) Choose the height of the cliff where the rock is thrown from the origin of the coordinate system. Although it is arbitrary, we typically choose time t = 0 to represent to the origin. (b) The equation that describes the horizontal motion is [latex] x={x}_{0}+{five}_{x}t. [/latex] With [latex] {ten}_{0}=0, [/latex] this equation becomes [latex] x={5}_{x}t. [/latex] (c) (Figure) through (Figure) and (Figure) describe the vertical motion, but since [latex] {y}_{0}=0\,\text{and}\,{five}_{0y}=0, [/latex] these equations simplify profoundly to become [latex] y=\frac{i}{ii}({five}_{0y}+{5}_{y})t=\frac{1}{two}{5}_{y}t,\enspace [/latex][latex] {v}_{y}=\text{−}gt,\enspace [/latex][latex] y=-\frac{i}{2}g{t}^{2},\enspace [/latex] and [latex] {v}_{y}^{2}=-2gy. [/latex] (d) We employ the kinematic equations to find the ten and y components of the velocity at the point of impact. Using [latex] {five}_{y}^{2}=-2gy [/latex] and noting the point of touch on is −100.0 m, nosotros observe the y component of the velocity at impact is [latex] {five}_{y}=44.three\,\text{grand}\text{/}\text{s}. [/latex] We are given the x component, [latex] {v}_{10}=15.0\,\text{m}\text{/}\text{s}, [/latex] so nosotros can calculate the total velocity at impact: five = 46.8 chiliad/southward and [latex] \theta =71.3\text{°} [/latex] beneath the horizontal.

Case

Computing Projectile Move: Lawn tennis Player

A tennis thespian wins a match at Arthur Ashe stadium and hits a ball into the stands at thirty grand/south and at an angle [latex] 45\text{°} [/latex] higher up the horizontal ((Effigy)). On its way down, the ball is caught by a spectator 10 k to a higher place the signal where the ball was hit. (a) Summate the time it takes the lawn tennis ball to reach the spectator. (b) What are the magnitude and management of the ball's velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball.

Effigy iv.14 The trajectory of a tennis ball hit into the stands.

Strategy

Again, resolving this ii-dimensional motility into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical movement solitary. Thus, we solve for t beginning. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the concluding velocity. Thus, nosotros recombine the vertical and horizontal results to obtain [latex] \overset{\to }{v} [/latex] at final time t, adamant in the kickoff part of the example.

Solution

(a) While the brawl is in the air, it rises and then falls to a concluding position 10.0 m higher than its starting altitude. We can detect the time for this by using (Figure):

[latex] y\,\text{=}\,{y}_{0}\,\text{+}\,{v}_{0y}t-\frac{i}{ii}m{t}^{2}. [/latex]

If we take the initial position [latex] {y}_{0} [/latex] to be zip, then the final position is y = 10 yard. The initial vertical velocity is the vertical component of the initial velocity:

[latex] {5}_{0y}={v}_{0}\text{sin}\,{\theta }_{0}=(xxx.0\,\text{1000}\text{/}\text{s})\text{sin}\,45\text{°}=21.ii\,\text{g}\text{/}\text{s}. [/latex]

Substituting into (Figure) for y gives the states

[latex] x.0\,\text{m}=(21.2\,\text{m/south})t-(4.90\,{\text{grand/s}}^{\text{2}}){t}^{2}. [/latex]

Rearranging terms gives a quadratic equation in t:

[latex] (iv.90\,{\text{m/south}}^{\text{ii}}){t}^{2}-(21.2\,\text{thou/s})t+10.0\,\text{m}=0. [/latex]

Apply of the quadratic formula yields t = iii.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way upward and in one case on the way downwardly—we take the longer solution for the time it takes the ball to reach the spectator:

[latex] t=iii.79\,\text{south}\text{.} [/latex]

The fourth dimension for projectile motion is determined completely by the vertical movement. Thus, whatever projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 g below its starting altitude spends 3.79 southward in the air.

(b) Nosotros can find the last horizontal and vertical velocities [latex] {v}_{x} [/latex] and [latex] {five}_{y} [/latex] with the use of the consequence from (a). So, nosotros can combine them to discover the magnitude of the total velocity vector [latex] \overset{\to }{5} [/latex] and the bending [latex] \theta [/latex] information technology makes with the horizontal. Since [latex] {five}_{x} [/latex] is abiding, we tin can solve for it at whatsoever horizontal location. We choose the starting bespeak because we know both the initial velocity and the initial bending. Therefore,

[latex] {v}_{ten}={v}_{0}\text{cos}{\theta }_{0}=(thirty\,\text{m}\text{/}\text{south})\text{cos}\,45\text{°}=21.2\,\text{grand}\text{/}\text{south}. [/latex]

The concluding vertical velocity is given by (Figure):

[latex] {five}_{y}={v}_{0y}-gt. [/latex]

Since [latex] {v}_{0y} [/latex] was found in office (a) to exist 21.two m/s, nosotros have

[latex] {v}_{y}=21.2\,\text{m}\text{/}\text{due south}-9.eight\,\text{one thousand}\text{/}{\text{s}}^{ii}(three.79\,\text{s})=-15.9\,\text{m}\text{/}\text{south}. [/latex]

The magnitude of the concluding velocity [latex] \overset{\to }{v} [/latex] is

[latex] v=\sqrt{{5}_{ten}^{2}+{v}_{y}^{ii}}=\sqrt{{(21.2\,\text{grand}\text{/}\text{s})}^{2}+{(\text{−}\,\text{fifteen}.9\,\text{one thousand}\text{/}\text{s})}^{2}}=26.5\,\text{m}\text{/}\text{s}. [/latex]

The management [latex] {\theta }_{v} [/latex] is plant using the inverse tangent:

[latex] {\theta }_{v}={\text{tan}}^{-i}(\frac{{v}_{y}}{{five}_{x}})={\text{tan}}^{-i}(\frac{21.two}{-fifteen.ix})=-53.i\text{°}. [/latex]

Significance

(a) As mentioned earlier, the time for projectile motion is determined completely past the vertical motility. Thus, whatsoever projectile that has an initial vertical velocity of 21.ii thousand/southward and lands 10.0 m below its starting distance spends iii.79 due south in the air. (b) The negative bending means the velocity is [latex] 53.1\text{°} [/latex] below the horizontal at the indicate of impact. This event is consistent with the fact that the ball is impacting at a indicate on the other side of the noon of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity nosotros expect since it is impacting 10.0 m above the launch top.

Fourth dimension of Flight, Trajectory, and Range

Of interest are the fourth dimension of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations requite useful expressions for these quantities, which are derived in the following sections.

Time of flying

We can solve for the time of flight of a projectile that is both launched and impacts on a apartment horizontal surface by performing some manipulations of the kinematic equations. We annotation the position and displacement in y must exist zero at launch and at touch on an even surface. Thus, nosotros set the displacement in y equal to zilch and find

[latex] y-{y}_{0}={five}_{0y}t-\frac{i}{ii}g{t}^{2}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}g{t}^{2}=0. [/latex]

Factoring, nosotros have

[latex] t({v}_{0}\text{sin}{\theta }_{0}-\frac{gt}{two})=0. [/latex]

Solving for t gives usa

[latex] {T}_{\text{tof}}=\frac{ii({5}_{0}\text{sin}{\theta }_{0})}{g}. [/latex]

This is the time of flying for a projectile both launched and impacting on a flat horizontal surface. (Figure) does not use when the projectile lands at a different meridian than it was launched, as nosotros saw in (Effigy) of the tennis player hit the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The fourth dimension of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity every bit on Earth would exist airborne six times every bit long.

Trajectory

The trajectory of a projectile can exist found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(ten). We take [latex] {x}_{0}={y}_{0}=0 [/latex] so the projectile is launched from the origin. The kinematic equation for x gives

[latex] x={v}_{0x}t⇒t=\frac{x}{{5}_{0x}}=\frac{x}{{v}_{0}\text{cos}{\theta }_{0}}. [/latex]

Substituting the expression for t into the equation for the position [latex] y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{i}{2}g{t}^{2} [/latex] gives

[latex] y=({v}_{0}\text{sin}\,{\theta }_{0})(\frac{10}{{v}_{0}\text{cos}\,{\theta }_{0}})-\frac{1}{ii}grand{(\frac{x}{{5}_{0}\text{cos}\,{\theta }_{0}})}^{two}. [/latex]

Rearranging terms, nosotros accept

[latex] y=(\text{tan}\,{\theta }_{0})x-[\frac{one thousand}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}. [/latex]

This trajectory equation is of the form [latex] y=ax+b{ten}^{2}, [/latex] which is an equation of a parabola with coefficients

[latex] a=\text{tan}\,{\theta }_{0},\enspaceb=-\frac{one thousand}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}. [/latex]

Range

From the trajectory equation nosotros tin also find the range, or the horizontal distance traveled by the projectile. Factoring (Effigy), we have

[latex] y=x[\text{tan}\,{\theta }_{0}-\frac{g}{ii{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}10]. [/latex]

The position y is nada for both the launch signal and the touch on point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions ten = 0, corresponding to the launch point, and

[latex] x=\frac{two{v}_{0}^{two}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{1000}, [/latex]

corresponding to the impact point. Using the trigonometric identity [latex] 2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}2\theta [/latex] and setting x = R for range, we detect

[latex] R=\frac{{5}_{0}^{2}\text{sin}2{\theta }_{0}}{g}. [/latex]

Annotation particularly that (Figure) is valid just for launch and impact on a horizontal surface. We encounter the range is directly proportional to the square of the initial speed [latex] {v}_{0} [/latex] and [latex] \text{sin}2{\theta }_{0} [/latex], and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Globe for the aforementioned initial velocity. Furthermore, we see from the factor [latex] \text{sin}2{\theta }_{0} [/latex] that the range is maximum at [latex] 45\text{°}. [/latex] These results are shown in (Figure). In (a) we run across that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at [latex] 45\text{°}. [/latex] This is true merely for weather neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the aforementioned range is found for two initial launch angles that sum to [latex] xc\text{°}. [/latex] The projectile launched with the smaller bending has a lower apex than the higher angle, but they both have the aforementioned range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle.

Figure iv.15 Trajectories of projectiles on level ground. (a) The greater the initial speed [latex] {five}_{0}, [/latex] the greater the range for a given initial bending. (b) The outcome of initial angle [latex] {\theta }_{0} [/latex] on the range of a projectile with a given initial speed. Notation that the range is the same for initial angles of [latex] fifteen\text{°} [/latex] and [latex] 75\text{°}, [/latex] although the maximum heights of those paths are unlike.

Instance

Comparing Golf game Shots

A golfer finds himself in two unlike situations on unlike holes. On the second hole he is 120 m from the green and wants to hit the brawl 90 one thousand and permit it run onto the green. He angles the shot low to the ground at [latex] thirty\text{°} [/latex] to the horizontal to allow the ball roll after touch on. On the 4th pigsty he is xc thou from the green and wants to allow the brawl drop with a minimum amount of rolling afterward bear on. Hither, he angles the shot at [latex] 70\text{°} [/latex] to the horizontal to minimize rolling after impact. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the ball at the 2d hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

Strategy

We see that the range equation has the initial speed and angle, so nosotros tin can solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

Solution

(a) [latex] R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}⇒{five}_{0}=\sqrt{\frac{Rg}{\text{sin}\,two{\theta }_{0}}}=\sqrt{\frac{ninety.0\,\text{yard}(9.viii\,\text{one thousand}\text{/}{\text{s}}^{2})}{\text{sin}(ii(lxx\text{°}))}}=37.0\,\text{m}\text{/}\text{due south} [/latex]

(b) [latex] R=\frac{{v}_{0}^{two}\text{sin}\,two{\theta }_{0}}{k}⇒{v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(ix.8\,\text{g}\text{/}{\text{s}}^{two})}{\text{sin}(two(thirty\text{°}))}}=31.ix\,\text{g}\text{/}\text{s} [/latex]

(c)

[latex] \brainstorm{assortment}{cc} y=10[\text{tan}\,{\theta }_{0}-\frac{one thousand}{2{({five}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{Second hole:}\,y=x[\text{tan}\,70\text{°}-\frac{nine.8\,\text{m}\text{/}{\text{southward}}^{two}}{2{[(37.0\,\text{yard}\text{/}\text{due south)(}\text{cos}\,70\text{°})]}^{2}}x]=2.75x-0.0306{ten}^{2}\hfill \\ \text{Fourth pigsty:}\,y=x[\text{tan}\,30\text{°}-\frac{9.8\,\text{m}\text{/}{\text{south}}^{ii}}{ii{[(31.ix\,\text{m}\text{/}\text{s)(}\text{cos}thirty\text{°})]}^{two}}x]=0.58x-0.0064{x}^{2}\hfill \end{array} [/latex]

(d) Using a graphing utility, nosotros tin compare the two trajectories, which are shown in (Effigy).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees.

Effigy 4.16 Ii trajectories of a golf ball with a range of 90 m. The impact points of both are at the same level as the launch betoken.

Significance

The initial speed for the shot at [latex] lxx\text{°} [/latex] is greater than the initial speed of the shot at [latex] 30\text{°}. [/latex] Annotation from (Figure) that ii projectiles launched at the same speed but at different angles have the same range if the launch angles add together to [latex] 90\text{°}. [/latex] The launch angles in this example add to requite a number greater than [latex] ninety\text{°}. [/latex] Thus, the shot at [latex] lxx\text{°} [/latex] has to have a greater launch speed to accomplish xc thousand, otherwise information technology would land at a shorter distance.

Cheque Your Understanding

If the two golf shots in (Figure) were launched at the same speed, which shot would have the greatest range?

The golf game shot at [latex] 30\text{°}. [/latex]

When we speak of the range of a projectile on level basis, nosotros assume R is very small compared with the circumference of Globe. If, still, the range is large, Earth curves away beneath the projectile and the dispatch resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given before considering the projectile has farther to fall than it would on level ground, every bit shown in (Figure), which is based on a cartoon in Newton's Principia . If the initial speed is great plenty, the projectile goes into orbit. Globe'southward surface drops 5 thousand every 8000 m. In 1 due south an object falls v m without air resistance. Thus, if an object is given a horizontal velocity of [latex] 8000\,\text{g}\text{/}\text{s} [/latex] (or [latex] 18,000\text{mi}\text{/}\text{hour}) [/latex] near Earth'southward surface, it volition go into orbit around the planet considering the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a depression Globe orbit when it was operational, or whatever satellite in a depression World orbit. These and other aspects of orbital motion, such as Globe's rotation, are covered in greater depth in Gravitation.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.

Figure four.17 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avert air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because World curves away below its path. With a speed of 8000 g/s, orbit is achieved.

Summary

  • Projectile motion is the movement of an object subject only to the acceleration of gravity, where the acceleration is constant, as near the surface of Earth.
  • To solve projectile motion bug, we analyze the move of the projectile in the horizontal and vertical directions using the one-dimensional kinematic equations for ten and y.
  • The time of flight of a projectile launched with initial vertical velocity [latex] {v}_{0y} [/latex] on an even surface is given past

    [latex] {T}_{tof}=\frac{2({v}_{0}\text{sin}\,\theta )}{k}. [/latex]

    This equation is valid simply when the projectile lands at the same elevation from which information technology was launched.

  • The maximum horizontal distance traveled past a projectile is called the range. Once again, the equation for range is valid only when the projectile lands at the aforementioned height from which it was launched.

Conceptual Questions

Answer the post-obit questions for projectile motion on level basis assuming negligible air resistance, with the initial angle being neither [latex] 0\text{°} [/latex] nor [latex] xc\text{°}: [/latex] (a) Is the velocity always zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity always exist the same as the initial velocity at a time other than at t = 0? (d) Tin can the speed ever be the same equally the initial speed at a time other than at t = 0?

a. no; b. minimum at apex of trajectory and maximum at launch and bear on; c. no, velocity is a vector; d. yeah, where it lands

Respond the following questions for projectile motion on level ground bold negligible air resistance, with the initial bending existence neither [latex] 0\text{°} [/latex] nor [latex] 90\text{°}: [/latex] (a) Is the dispatch ever zero? (b) Is the dispatch ever in the same management equally a component of velocity? (c) Is the acceleration e'er opposite in direction to a component of velocity?

A dime is placed at the edge of a table and so it hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime head on. Which money hits the footing first?

They both striking the basis at the aforementioned fourth dimension.

Problems

A bullet is shot horizontally from shoulder elevation (1.5 m) with and initial speed 200 m/s. (a) How much fourth dimension elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

a. [latex] t=0.55\,\text{south} [/latex], b. [latex] ten=110\,\text{m} [/latex]

A marble rolls off a tabletop 1.0 m high and hits the floor at a point 3.0 grand abroad from the table's border in the horizontal management. (a) How long is the marble in the air? (b) What is the speed of the marble when it leaves the table'due south edge? (c) What is its speed when it hits the floor?

A sprint is thrown horizontally at a speed of ten m/s at the bull's-eye of a dartboard ii.4 m away, as in the following figure. (a) How far below the intended target does the dart hit? (b) What does your respond tell you about how proficient dart players throw their darts?

An airplane flying horizontally with a speed of 500 km/h at a height of 800 yard drops a crate of supplies (see the following effigy). If the parachute fails to open up, how far in front end of the release point does the crate hit the ground?

An airplane releases a package. The airplane has a horizontal velocity of 500 kilometers per hour. The package's trajectory is the right half of a downward-opening parabola, initially horizontal at the airplane and curving down until it hits the ground.

Suppose the airplane in the preceding problem fires a projectile horizontally in its direction of motility at a speed of 300 m/s relative to the plane. (a) How far in front of the release point does the projectile hit the footing? (b) What is its speed when it hits the ground?

a., [latex] t=12.viii\,\text{s,}\enspacex=5619\,\text{m} [/latex] b. [latex] {5}_{y}=125.0\,\text{m}\text{/}\text{s,}\enspace{five}_{x}=439.0\,\text{m}\text{/}\text{s,}\enspace|\overset{\to }{v}|=456.0\,\text{thou}\text{/}\text{s} [/latex]

A fastball pitcher can throw a baseball at a speed of 40 m/s (90 mi/h). (a) Bold the bullpen tin can release the ball 16.7 one thousand from home plate so the brawl is moving horizontally, how long does it take the ball to reach home plate? (b) How far does the ball drop betwixt the pitcher'south manus and home plate?

A projectile is launched at an angle of [latex] 30\text{°} [/latex] and lands twenty southward later at the same top as it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Summate the deportation from the point of launch to the position on its trajectory at 15 due south.

A basketball histrion shoots toward a handbasket six.1 yard away and 3.0 m higher up the floor. If the brawl is released 1.8 m higher up the floor at an angle of [latex] lx\text{°} [/latex] in a higher place the horizontal, what must the initial speed be if information technology were to go through the basket?

At a particular instant, a hot air balloon is 100 thou in the air and descending at a constant speed of ii.0 1000/s. At this exact instant, a girl throws a brawl horizontally, relative to herself, with an initial speed of 20 m/s. When she lands, where will she find the brawl? Ignore air resistance.

[latex] -100\,\text{m}=(-2.0\,\text{m}\text{/}\text{southward})t-(4.9\,\text{thousand}\text{/}{\text{s}}^{two}){t}^{two}, [/latex][latex] t=four.three\,\text{due south,} [/latex][latex] ten=86.0\,\text{thousand} [/latex]

A man on a motorbike traveling at a uniform speed of x chiliad/south throws an empty tin straight upward relative to himself with an initial speed of 3.0 k/due south. Notice the equation of the trajectory as seen by a police officer on the side of the route. Assume the initial position of the can is the indicate where it is thrown. Ignore air resistance.

An athlete can jump a distance of eight.0 m in the broad jump. What is the maximum distance the athlete can jump on the Moon, where the gravitational acceleration is one-6th that of World?

[latex] {R}_{Moon}=48\,\text{g} [/latex]

The maximum horizontal distance a boy tin can throw a ball is 50 k. Assume he can throw with the aforementioned initial speed at all angles. How high does he throw the ball when he throws it straight upward?

A rock is thrown off a cliff at an angle of [latex] 53\text{°} [/latex] with respect to the horizontal. The cliff is 100 k high. The initial speed of the rock is 30 m/s. (a) How high higher up the edge of the cliff does the rock rise? (b) How far has it moved horizontally when it is at maximum altitude? (c) How long after the release does it hit the ground? (d) What is the range of the rock? (e) What are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 s, t = 4.0 southward, and t = 6.0 s?

Trying to escape his pursuers, a secret amanuensis skis off a slope inclined at [latex] xxx\text{°} [/latex] below the horizontal at 60 km/h. To survive and state on the snow 100 grand beneath, he must clear a gorge sixty thou wide. Does he make it? Ignore air resistance.

A skier is moving with velocity v sub 0 down a slope that is inclined at 30 degrees to the horizontal. The skier is at the edge of a 60 m wide gap. The other side of the gap is 100 m lower.

A golfer on a fairway is 70 m abroad from the green, which sits below the level of the fairway past xx m. If the golfer hits the ball at an bending of [latex] 40\text{°} [/latex] with an initial speed of 20 m/s, how shut to the green does she come up?

A projectile is shot at a hill, the base of which is 300 one thousand away. The projectile is shot at [latex] threescore\text{°} [/latex] above the horizontal with an initial speed of 75 m/s. The hill tin can be approximated past a plane sloped at [latex] 20\text{°} [/latex] to the horizontal. Relative to the coordinate arrangement shown in the post-obit figure, the equation of this straight line is [latex] y=(\text{tan}20\text{°})x-109. [/latex] Where on the hill does the projectile state?

A projectile is shot from the origin at a hill, the base of which is 300 m away. The projectile is shot at 60 degrees above the horizontal with an initial speed of 75 m/s. The hill is sloped away from the origin at 20 degrees to the horizontal. The slope is expressed as the equation y equals (tan of 20 degrees) times x minus 109.

An astronaut on Mars kicks a soccer ball at an angle of [latex] 45\text{°} [/latex] with an initial velocity of 15 k/s. If the dispatch of gravity on Mars is three.vii m/s, (a) what is the range of the soccer kicking on a apartment surface? (b) What would be the range of the same kick on the Moon, where gravity is one-6th that of Earth?

Mike Powell holds the record for the long jump of 8.95 thousand, established in 1991. If he left the ground at an bending of [latex] 15\text{°}, [/latex] what was his initial speed?

MIT'southward robot cheetah can jump over obstacles 46 cm high and has speed of 12.0 km/h. (a) If the robot launches itself at an bending of [latex] threescore\text{°} [/latex] at this speed, what is its maximum height? (b) What would the launch angle have to be to reach a elevation of 46 cm?

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an bending of [latex] 40\text{°} [/latex] with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flying?

Drew Brees of the New Orleans Saints can throw a football 23.0 m/s (50 mph). If he angles the throw at [latex] x\text{°} [/latex] from the horizontal, what altitude does it go if information technology is to be caught at the same elevation as it was thrown?

[latex] R=18.5\,\text{chiliad} [/latex]

The Lunar Roving Vehicle used in NASA's late Apollo missions reached an unofficial lunar land speed of 5.0 1000/s by astronaut Eugene Cernan. If the rover was moving at this speed on a apartment lunar surface and striking a small-scale bump that projected information technology off the surface at an angle of [latex] 20\text{°}, [/latex] how long would it be "airborne" on the Moon?

A soccer goal is 2.44 m high. A player kicks the ball at a distance 10 m from the goal at an angle of [latex] 25\text{°}. [/latex] What is the initial speed of the soccer brawl?

[latex] y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{two}⇒{v}_{0}=16.4\,\text{yard}\text{/}\text{s} [/latex]

Olympus Mons on Mars is the largest volcano in the solar organization, at a elevation of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you take to fire a projectile from a cannon horizontally to clear the volcano and country on the surface of Mars? Note that Mars has an acceleration of gravity of [latex] 3.7\,\text{m}\text{/}{\text{s}}^{ii}. [/latex]

In 1999, Robbie Knievel was the first to leap the 1000 Coulee on a motorcycle. At a narrow part of the canyon (69.0 k broad) and traveling 35.eight m/due south off the takeoff ramp, he reached the other side. What was his launch angle?

[latex] R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{m}⇒{\theta }_{0}=xv.0\text{°} [/latex]

Y'all throw a baseball at an initial speed of fifteen.0 thousand/s at an angle of [latex] 30\text{°} [/latex] with respect to the horizontal. What would the ball's initial speed have to be at [latex] 30\text{°} [/latex] on a planet that has twice the acceleration of gravity every bit Globe to attain the same range? Consider launch and impact on a horizontal surface.

Aaron Rogers throws a football at twenty.0 m/due south to his wide receiver, who runs straight down the field at 9.4 m/s for 20.0 m. If Aaron throws the football when the broad receiver has reached 10.0 grand, what angle does Aaron have to launch the ball so the receiver catches it at the 20.0 thou marking?

Glossary

projectile motion
motion of an object subject simply to the acceleration of gravity
range
maximum horizontal distance a projectile travels
time of flight
elapsed time a projectile is in the air
trajectory
path of a projectile through the air

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Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-3-projectile-motion/

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