An inelastic ball is projetced with a velocity `'u'` at an angle `'alpha'` to the horizontal, towards a wall distant `'d'` from the point of projection. After collision the ball returns to the point of projection (Co-efficient of restitution between sphere and wall is `'e'`)

The horizontal distance `'d'` from the wall is

Options

A. `(u^(2)sin 2 alpha)/(g)((e)/(1+e))`
B. `(u^(2)sin alpha)/(g)((e)/(1+e))`
C. `(u^(2)cos 2alpha)/(g)((e)/(1+e))`
D. `(u^(2)sin 2alpha)/(2g)((e)/(1+e))`

Answer

A

Explanation

Time taken ro reach the wall`(t_(1)) ~~ (d)/(u cos alpha)`
Time taken to return to the point of projection after impact `(t_(2)) = (d)/(eu cos alpha)`.
Total time of flight `(T) = t_(1)+t_(2) = (d)/(u cos alpha)((1+e)/(e))`
There is no change in the vertical component of the velocity after impact
`therefore T = (2u sin alpha)/(g)` .........(1)
`(d)/(u cos alpha)((1+e)/(e)) = (2u sin alpha)/(g)`
`d = (u^(2) sin 2alpha)/(g) ((e)/(1+e)).........(2)`
From Fig. `Tantheta = (y)/(d) , y = d tan theta`
`y = (tan alpha)x-(1)/(2)(g)/(u^(2) cos^(2)alpha) xx x^(2)`
`(tan theta)d = (tan alpha)d-(1)/(2)(g)/(u^(2)cos^(2) alpha)xx d^(2)`
`tan theta = tan alpha - (1)/(2)(g)/(u^(2) cos^(2) alpha) xx d` .......(3)
From (2) and (3)
`tan theta = (tan alpha)/(1+e)`