
Projectiles move horizontally at uniform velocity and vertically, accelerated by gravity. The motions can be analyzed separately, and this is the key to solving problems, such as a projectile’s range or time of flight. Motion along a ramp is accelerated by gravity if the ramp is inclined: the steeper the angle, the greater the acceleration. Rolling friction reduces this acceleration, but is less significant as the ramp steepens.

trajectory, projectile, range, inclined plane, ramp coordinates


$$x={v}_{x0}t$$
$${v}_{x}={v}_{x0}$$

 $$y={v}_{y0}t{\scriptscriptstyle \frac{1}{2}}g{t}^{2}$$
$${v}_{y}={v}_{y0}gt$$


$$x={v}_{0}t\text{\u202f}\mathrm{cos}\text{\u202f}\theta $$
$${v}_{x}={v}_{0}\text{\u202f}\mathrm{cos}\text{\u202f}\theta $$

 $$y={v}_{0}t\text{\u202f}\mathrm{sin}\text{\u202f}\theta {\scriptscriptstyle \frac{1}{2}}g{t}^{2}$$
$${v}_{y}={v}_{0}\text{\u202f}\mathrm{sin}\text{\u202f}\theta gt$$

 $$x=\frac{2{v}_{0}^{2}\text{\u202f}\mathrm{sin}\theta \text{\u202f}\mathrm{cos}\theta}{g}$$ 

$${a}_{ramp}=\left(\frac{h}{L}\right)g$$

 $${a}_{x}=g\mathrm{sin}\theta $$

 $${a}_{x}=g(\mathrm{sin}\theta {\mu}_{r}\mathrm{cos}\theta )$$



Review problems and questions 


A soccer ball is kicked and leaves the ground with an initial
speed v_{0} at an angle θ. Assume that the initial speed is fixed but the angle can vary. Match each quantity in the lefthand column with one quantity in the righthand column:
time ball is in air  largest when θ = 0° 
horizontal speed  largest when 0 < θ < 90° 
horizontal distance  largest when θ = 90° 

time ball is in air  →  largest when θ = 90°  horizontal speed  →  largest when θ = 0°  horizontal distance  →  largest when 0 < θ < 90° 
When θ = 90°, the entire launch velocity is in the vertical (y) direction. The acceleration due to gravity subtracts 9.8 m/s each second from the initial vertical velocity. The ball reaches its greatest height when its vertical velocity is reduced to zero—and that is half the travel time. Therefore the ball stays aloft the longest when θ = 90°.
In contrast, when θ = 0° the entire launch velocity is in the horizontal (x) direction. This means that the horizontal speed is maximized when θ = 0°.
Finally, the horizontal distance (or range) is the product of the time in flight and the horizontal speed. The ball has zero range if it goes straight up (θ = 90°). It never leaves the ground, however, if θ = 0°. Therefore the range is maximized by an angle between 0° and 90°. (In fact, the range is greatest when θ = 45°.)


A soccer ball is kicked at an upward angle θ between 0° and 90°. Its initial vertical (y) velocity v_{y}_{0} is 19.6 m/s. It follows a parabolic trajectory and lands at the end of its range.
 How long (how much time) is the ball in the air?
 How high does the ball get?
 If the ball’s range is 60 m, is the angle θ less than, equal to, or greater than 45°?

Answer:  The ball is in the air for a total of 4 s.
 The ball reaches a maximum height of h = 19.6 m.
 The ball lifts off at an angle θ that is greater than 45°.
Solution: The ball is in the air for a total of 4 s. You can solve this problem in several ways. One way is to find out how long it takes for the ball to stop rising (that is, for v_{y}_{0} to go to zero); double that time for the total time in flight. Another way is to solve for the time at which the ball’s height, or y coordinate, returns to zero.
Asked: You are asked to find the time the ball is in the air. Given: initial vertical velocity v_{y}_{0} = 19.6 m/s Relationship: v_{y} = v_{y}_{0} − gt Solution: Solve for the time t that the ball takes to reach the top of its trajectory (where v_{y} = 0). Starting with:$${v}_{y}={v}_{y0}gt$$we solve for t:$$\begin{array}{ccc}gt={v}_{y0}{v}_{y}& \Rightarrow & t=\frac{{v}_{y0}{v}_{y}}{g}=\frac{19.6\text{m/s}0}{9.8{\text{m/s}}^{2}}=2\text{s}\end{array}$$The ball takes 2 s to rise to its maximum height. By symmetry, it then takes an equal amount of time to fall back to the ground. The total time in flight is therefore 4 s.  The ball reaches a maximum height of h = 19.6 meters. This can be calculated using the formula for a projectile’s ycoordinate:$$y={y}_{0}+{v}_{y0}t{\scriptscriptstyle \frac{1}{2}}g{t}^{2}=\left(0\text{m}\right)+\left(19.6\text{m/s}\right)\left(2\text{s}\right){\scriptscriptstyle \frac{1}{2}}\left(9.8{\text{m/s}}^{2}\right){\left(2\text{s}\right)}^{2}=19.8\text{m}$$
 The ball lifts off at an angle θ that is greater than 45°. Ignoring air resistance, a projectile maintains a constant horizontal (x) velocity until it lands or strikes an obstacle. Since we know the range (60 m) and the time of flight (4 s), we can immediately calculate the initial horizontal velocity:$${v}_{\mathrm{x}0}=\frac{60\text{m}}{4\text{s}}=15\text{m}/\text{s}$$Since the initial vertical speed (19.6 m/s) is greater than this horizontal speed, the ball must have left the ground at an angle greater than 45°. (Think of a right triangle whose height is greater than its base.)


A small toy car is placed upon a ramp that is 2 m long. The car is released (with zero initial velocity) from the top of the ramp. Assume that friction is negligible.
 How high is the top if the car takes 2.0 s to reach the bottom?
 Now suppose that you want the car to go from top to bottom in just one second (and you are still releasing it, not pushing it, so its initial velocity remains zero). In other words, you want to divide the travel time by two. Will you achieve this by doubling the height, or will you have to raise the ramp by a different amount?

Answer:  The ramp height must be h = 0.20 m.
 You need to more than double the ramp height to halve the travel time.
Solution: The ramp height must be h = 0.20 meters if the car is to take t = 2.0 seconds to travel from top to bottom.
Asked: height h of the top of the ramp Given: ramp length L = 2.0 m; travel time t = 2.0 s Relationships: acceleration along ramp axis a_{ramp} = (h/L)g; distance traveled by a uniformly accelerating object (with zero initial velocity) d = (½)at^{2} Solution: Use the ramp acceleration equation to solve for the desired height h:$$\begin{array}{ccc}{a}_{ramp}=\frac{h}{L}g& \Rightarrow & h=\frac{L}{g}{a}_{ramp}\end{array}$$In this equation we don’t know a_{ramp}, so we must find it using another equation. Use the equation of motion, where the distance d is the length of the ramp L, and solve for the unknown acceleration along the ramp:$${\begin{array}{ccc}L={\scriptscriptstyle \frac{1}{2}}{a}_{ramp}{t}^{2}& \Rightarrow & a\end{array}}_{ramp}=\frac{2L}{{t}^{2}}$$Now substitute this expression back into the equation for height and calculate the answer:$$h=\frac{L}{g}{a}_{ramp}=\frac{L}{g}\frac{2L}{{t}^{2}}=\frac{2{L}^{2}}{g{t}^{2}}=\frac{2{\left(2.0\text{m}\right)}^{2}}{\left(9.8{\text{m/s}}^{2}\right){\left(2.0\text{s}\right)}^{2}}=0.20\text{m}$$  You need to more than double the ramp height to halve the travel time. In the solution to part a above, height is inversely proportional to the square of the time. This means that half the time corresponds to four times the height, or 0.8 m.

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