If at a height of ` 40 m`, the direction of moton of a projectile makes an angle `pi //4` with the orizontal, then its initial velocity and angle of projection are, respectively,

Options

A. ` 30 , 1/2 cos^(-1) (- 4/5)`
B. ` 30 1/2 cos^(-1) (-1/2)`
C. ` 50 1/2 cos^(1) (- 8 /(25)`
D. 60 1/2 cos^(-1) (- 1/4)`

Answer

C

Explanation

Let (u) be the initial velocity of projectile and ` theta` be the angel of projmection. Taking vertical upwarawd motion for hieght (h) , we have
` u =u sin theta, a=- g, S=h, v=v-y, As
` v^2 = u^2 + 2 aS,
so ` v_y^2 = u^2 sin^2 theta- 2 gh` ...(i)
At height ` h, tahn (pi)/4 = v_y/v_x ` or ` v_y =v-x = u cos theta`
:. ` u^2 cos^2 theta = u ^2 sin^2 sin^2 theta - 2 gt`
or ` u^2 (cos^2 theta- sin ^2 theta)=- 2 gt` ltBrgt or ` u^2 cos 2 theta =- 2 gt ` ....(ii)
The equation (ii) is valid only is ` u= 50 m//s`
and ` cos ` 2 theta =- 8//25`
or ` theta 1/2 cos ^(-1) (- 8//25)`.