Question

A small ball thrown at an initial velocity ${\nu }_0$ at an angle $\alpha$ to the horizontal strikest a vertical wall moving towards it at a horizontal velocity $\nu$ and is bounced to the point from which it was thrown.
Determine the time $t$ from the beginning of motion to the moment of impact, neglecting friction losses.

Answer 1

Since the wall is smooth,the impact against the wall does not alter the vertical component of the ball velocity. Therefore, the total time $t_1$ of motion of the ball is the total time of the ascent $t$ and descent of the body thrown upwards at a velocity ${\nu }_0\sin \alpha$ in the gravitational field. Consequently $t_1=2{\nu }_0\sin \alpha /g$ The motion of the ball along the horizontal is the sum of two motions. Before the collision with the wall, it moves at a velocity ${\nu }_0\cos \alpha$ After the collision, it traverses the same distance backwards,but at a different velocity. In order to calculate the velocity of the backward motion of the ball, it should be noted that the velocity at which the ball approaches the wall (along the horizontal ) is ${\nu }_0\cos \alpha +\nu$ Since the impact is perfectly elastic, the ball moves away from the wall after the collision at a velocity ${\nu }_0\cos \alpha +\nu$. Therefore, the ball has the following horizontal velocity relative to the ground:
$({\nu }_0\cos \alpha +\nu )+\nu ={\nu }_0\cos \alpha +2\nu$
If the time of motion before the impact is $t$, by equating the distances covered by the ball before and after the collision, we obtain the following equation:
${\nu }_0\cos \alpha .t=(t_1-t)({\nu }_0\cos \alpha +2\nu )$
Since the total time of motion of the ball is $t_1=2{\nu }_0\sin \alpha /g$ we find that
$t=\frac{{\nu }_0\sin \alpha ({\nu }_0\cos \alpha +2\nu )}{g({\nu }_0\cos \alpha +\nu )}$