Question

A ball is thrown with an initial horizontal velocity of 30 m/s from a height of 3m above the ground and 40m from a vertical wall. How far from the wall will the ball rebound for the first time? 

Answer 1

$v_x = 30m/s\\ h = 3m\\ x= 40m\\$

$v² = u²+ 2as\\ v_x² = u_x² + 2a_xx\\ v_² = 30² + 0\\ v_x = 30m/s$

$t = \dfrac{x}{v_y}= \dfrac{40}{30} =1.33s$

$y = u_yt+ \frac{1}{2} a_yt²\\ y = 0(1.33) + \frac{1}{2}(10)(1.33)²\\ y = 8.89m$

The ball would have fallen before it got to the wall.