A cannon is fixed with a smooth massive trolley car at an angle `theta` as shown in the figure.The trolley car slides from rest down the inclined plane of angle of inclination `beta`.
The muzzle velocity of the shell fired at `t=y_(0)` from the cannon is `u`,such that the shell moves perpendicular to the inclined just after the firing.
the difference in range of the shell relative to the trolley car and ground is:

Options

A. `(u^(2)sin 2theta)/(g cos beta)`
B. `(u^(2)cos theta)/(2g sin beta)`
C. `(u^(2) sin theta sin beta)/(2g)`
D. `(2U^(2) sin theta cos (theta-beta))/(g cos^(2) g)`

Answer

D

Explanation

Distance moved by shell `=(V_(0)cos theta)t+1/2g sin betat^(2)`
`t=(2U sin theta)/(g cos beta)` on solving `x=(2U^(2) sin theta)/(g cos^(2) beta)cos(theta-beta)`