A cannon on a level plane is aimed at an angle `theta` above the horizontal and a shell is fired with a muzzle velocity `v_(0)` towards a vertical cliff a distance D away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

Options

A. `D sin theta - (g D^2)/(2v_0^2 sin^2 theta)`
B. `Dcos theta - (g D^2)/(2v_0^2 cos^2 theta)`
C. `D tan theta - (gD^2)/(2v_0^2 cos^2 theta)`
D. `D tan theta - (gD^2)/(2v_0^2 sin^2 theta)`

Answer

C

Explanation

From the resultant path of the particle , when it is projected at angle `theta` with its velocity u is
`y=x tan theta -1/2 (gx^2)/(u^2 cos^2 theta)`

Where y denotes the instantaneous height of particle when it travels an instantaneous horizontal distance x. here x=D , `u=v_0`
so `y=D tan theta -1/2 (gD^2)/(v_0^2 cos^2 theta)`