A projectile is fired with a speed u at an angle `theta` above the horizontal field. The coefficient of restitution between the projectile and field is e. Find the position from the starting point when the projectile will land at its second collision

Options

A. `(e^(2)u^(2)sin 2theta)/g`
B. `((1-e^(2))u^(2)sin 2theta)/g`
C. `((1-e)u^(2)sinthetacostheta)/g`
D. `((1+e)u^(2)sin 2theta)/g`

Answer

D

Explanation

Vertical velocity after collision =`"e u" sintheta`
`"So " x=R+R'=("u"^(2)sin2theta)/g+(2ucostheta("eu "sin theta))/g`
`=("u"^(2)sin2theta)/(g)(1+e)`