Question

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.)

Answer 1

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

Answer 2

I'm getting 10.86 m for part a which isn't really reasonable.