The angle of which the velocity vector of a projectile thrown with a velocity u at an angle `theta` to the horizontal will take with the horizontal after time t of its being thrown up is

Options

A. `theta`
B. `tan^(-1)((theta)/(t))`
C. `tan^(-1)((vcos theta)/(v sintheta-g t))`
D. `tan^(-1)((v sin theta-g t)/(v cos theta))`

Answer

D

Explanation

`tan beta =(v_(y))/(v_(x))=(v sin theta-gt)/(v cos theta)`