Question

A ball is thrown with a purely horizontal velocity of 25m/s at a wall which is 15m away. The ball is thrown at a height of 1.6 m above the horizontal terrain. The ball has a coefficient of restitution of 0.75. After the ball impacts the wall and bounces off the wall back toward the thrower, how far from the wall will the ball initially strike the ground?

Answer 1

Given
A ball is thrown with a horizontal velocity (u2i) = 25 m/s towards the wall is 15 m away
Height of the ball from where it was thrown is h = 1.6 m

coefficient of restitution is = 0.75

By definition coefficient of restitution is The ratio of relative velocity after collision to the relative velocity before collision

   0.75 = (v1f-v2f)/(u1i -u2i)

here v1f and u1i refers to the wall which is stationary before and after collision

   0.75 = (0-v2f)/(0-(25))

   v2f = 18.75 m/s

with this velocity the ball will move from the wall

from horizontal projectile the range,

   Range R = u sqrt(2h/g)

here u is the velocity of the ball ,u = v2f = 18.75 m/s

  
   R = 18.75 sqrt(2*1.6/9.8) m

   R = 10.71 m

the wall will the ball initially strike the ground is 10.71 m

so some amount of energy has been lost