A squash ball typically rebounds from a surface with 25% of the speed with which it initially struck the surface. Suppose a squash ball is served in a shallow trajectory, from a height above the floor of 39 cm, at a launch angle of 6.0° above the horizontal, and at a distance of 11 m from the front wall.
(a) If it strikes the front wall exactly at the top of its parabolic trajectory, determine how high above the surface the ball strikes the wall?
(b) How far horizontally from the wall does it strike the floor, after rebounding? (Ignore any effects due to air resistance.)
s is initial speed
Vi = s sin 6
v = Vi - g t
at top v = 0
t = s sin 6 / 9.81
u = s cos 6
11 = u t
t = 11/(s cos 6)
so
11/(s cos 6) = s sin 6/9.81
solve for s, the initial speed
get t
then
h = .39 + Vi t - 4.9 t^2
for part b
u = s cos 6
new u = .25 s cos 6
then t to fall from
h = 4.9 t^2
use that t and new u to get distance from wall
I'm getting 10.86 m for part a which isn't really reasonable.
To solve these problems, we can use the principles of projectile motion and the given information.
(a) To determine how high above the surface the ball strikes the wall, we need to find the maximum height reached by the ball.
Step 1: Analyze the initial conditions
- The initial height (above the floor) is given as 39 cm, which is equivalent to 0.39 m.
- The launch angle is given as 6.0° above the horizontal.
- The distance from the front wall is given as 11 m.
Step 2: Calculate the initial velocity components
We need to find the magnitude of the initial velocity and its horizontal and vertical components.
- The horizontal component of the initial velocity (Vx) is given by Vx = V * cos(θ), where V is the magnitude of the initial velocity and θ is the launch angle.
- The vertical component of the initial velocity (Vy) is given by Vy = V * sin(θ).
Step 3: Find the time of flight
The time of flight (T) is the time it takes for the ball to reach its maximum height and then return to the same vertical position.
- The time to reach the maximum height is given by the formula T = Vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
- The total time of flight is double the time to reach the maximum height since the ball will take the same amount of time to come back down.
Step 4: Calculate the maximum height
The maximum height (H) can be determined using the formula H = Vy² / (2 * g).
Step 5: Determine the height above the surface
The height above the surface is the sum of the maximum height and the initial height.
(b) To find how far horizontally from the wall the ball strikes the floor, we need to determine the horizontal distance traveled by the ball after rebounding.
Step 1: Calculate the total horizontal distance traveled by the ball before rebounding.
The horizontal distance (D) can be calculated using the equation D = Vx * T.
Step 2: Determine the rebound horizontal distance.
Since the ball rebounds with 25% of its initial speed, we can calculate the rebound horizontal distance (DR) as follows: DR = D * (0.25).
With these steps, we can find the answers to parts (a) and (b) of the problem. Let's plug the values into the equations and calculate the results.