A small ball thrown at an initial velocity `v_(0)` at an angle `alpha` to the horizontal strikes elastically with a smooth vertical wall moving towards it at a horizontal velocity v and returns to the point from which it was thrown. Determine the time t from the beginning of motions to the moment of impact with vertical wall. Neglect friction losses.

Options

A. `t=(v_(0)sinalpha(v_(0)cosalpha+2v))/(g(v_(0)cosalpha+v))`
B. `t=(v_(0)sinalpha(v_(0)cosalpha-2v))/(g(v_(0)cosalpha+v))`
C. `t=(v_(0)sinalpha(v_(0)cosalpha+2v))/(g(v_(0)cosalpha-v))`
D. `t=(v_(0)cosalpha(v_(0)sinalpha+2v))/(g(v_(0)sinalpha+v))`

Answer

A

Explanation

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 and descent of the body thrown upwards at a velocity `v_(0) sinalpha` in the gravitational field. Consequently `t_(1)=2v_(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 `v_(0)cosalpha+v`. Since the impact is perfectly elastic, the ball moves away from the wall after the collision at a velocity `v_(0)cosalpha+v`. Therefore the ball has the following horizontal velocity relative tot he ground
`(v_(0)cosalpha+v)+v=v_(0)cosalpha+2v`.
If the time of motion before the impact is `t`. By equating the distance covered by the ball before ad after te collision, we obtain the following equation:
`v_(0)cosalpha alpha.t=(t_(1)-t)(v_(0)cosalpha+2v)`
since the total time of motion of the ball is `t_(1)=2v_(0)sin(alpha)/(g)`. we find the thhat
`t=(v_(0)sinalpha(v_(0)cosalpha+2v))/(g(v+(0)cosalpha+v))`