A cricket fielder can throw the cricket ball with a speed `v_(0)`.If the throws the ball while running with speed `u` at an angle `theta` to the horizontal.
The effective angle to the horizontal at which the ball is projected in air as seen by a spectator is

Options

A. `tan^(-1) [(v_(0)cos theta)/(u +v_(0) sin theta)]`
B. `tan^(-1) [(v_(0) sin theta)/(u+v_(0) cos theta)]`
C. `tan^(-)[(u)/(v_(0) cos theta + v_(0) sin theta)]`
D. `tan^(-1) [(v_(0) sin theta +v_(0) cos theta)/(u)]`

Answer

B

Explanation

Consider the adjacent diagram.

Initialy velocity in, x-direction, `u_(x) = u +v_(0) cos theta`
`u_(y) =` Initial velocity in y- direction `= v_(0) sin theta`
where angle of projection is `theta`.
Now, we can write
`tan theta = (u_(y))/(u_(x)) = (u_(0)sin theta)/(u +u_(0) cos theta) rArr theta = tan^(-1) ((v_(0)sin theta)/(u+ v_(0) cos theta))`