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A projectile is thrown upward with a vel...

A projectile is thrown upward with a velocity `v_(0)` at an angle `alpha` to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is

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))`

Text Solution

Verified by Experts

The correct Answer is:
A
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))`
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