A small ball thrown with initial velocity $v_{0}$ at an angle a to the horizon hit a vertical wall moving towards it with horizontally directed velocity $v(t)$ and bounced back to the point from which it was thrown. Determine at what time t after the throw did the ball collide collision of the ball with the wall? Neglect friction losses.
Decision:
Since the wall is smooth, the impact against the wall does not change the vertical component of the ball's velocity. Therefore, the total time of motion of the ball $t_{1}$ is the total time of ascent and descent to the initial height in the field of gravity of a body thrown upward with velocity $v_{0} \sin \alpha$. Hence, $t_{1} = 2v_{0} \sin \left ( \frac{ \alpha}{g} \right )$. The horizontal motion of the ball is made up of two parts of the path: before impact with the wall, it traveled with velocity $v_{0} \cos \alpha$; after impact, the ball traveled back the same path, but with a different velocity. To calculate the velocity of the ball's backward motion, note that the velocity of the ball and wall approaching each other (horizontally) was $v_{0} \cos \alpha + v$. Since the impact is absolutely elastic, after the impact the ball will move away from the wall with velocity $v_{0} \cos \alpha +v$, so relative to the ground it will have horizontal velocity
$(v_{0} \cos \alpha + v) + v = v_{0} \cos \alpha +2v$
If the ball flew for time t before hitting the wall, then, equating the paths traveled by it before and after the impact, we obtain the equation
$v_{0} \cos \alpha \cdot t=(t_{1}-t)(v_{0} \cos \alpha +2v)$
Hence, given that the total time of motion of the ball is $t_{1}=2v_{0} \sin \frac{ \alpha}{g}$, we obtain
$t=v_{0} \sin \frac{ \alpha(v_{0} \cos \alpha +2v)}{g(v_{0} \cos \alpha +v)}$
Decision:
Since the wall is smooth, the impact against the wall does not change the vertical component of the ball's velocity. Therefore, the total time of motion of the ball $t_{1}$ is the total time of ascent and descent to the initial height in the field of gravity of a body thrown upward with velocity $v_{0} \sin \alpha$. Hence, $t_{1} = 2v_{0} \sin \left ( \frac{ \alpha}{g} \right )$. The horizontal motion of the ball is made up of two parts of the path: before impact with the wall, it traveled with velocity $v_{0} \cos \alpha$; after impact, the ball traveled back the same path, but with a different velocity. To calculate the velocity of the ball's backward motion, note that the velocity of the ball and wall approaching each other (horizontally) was $v_{0} \cos \alpha + v$. Since the impact is absolutely elastic, after the impact the ball will move away from the wall with velocity $v_{0} \cos \alpha +v$, so relative to the ground it will have horizontal velocity
$(v_{0} \cos \alpha + v) + v = v_{0} \cos \alpha +2v$
If the ball flew for time t before hitting the wall, then, equating the paths traveled by it before and after the impact, we obtain the equation
$v_{0} \cos \alpha \cdot t=(t_{1}-t)(v_{0} \cos \alpha +2v)$
Hence, given that the total time of motion of the ball is $t_{1}=2v_{0} \sin \frac{ \alpha}{g}$, we obtain
$t=v_{0} \sin \frac{ \alpha(v_{0} \cos \alpha +2v)}{g(v_{0} \cos \alpha +v)}$