A particle is ejected from the tube at A with a velocity `V` at an angle `theta` with the verticle y-axis at a height h above the ground as shown. A strong horizontal wind gives the particle a constant horizontal acceleration a in the positive x-direction. If the particle strikes the ground at a point directly under its released position and the downward y-acceleration is taken as g then find h.

Options

A. `h=(2v^(2) sin theta cos theta)/(a)`
B. `h=(2v^(2) sin theta cos theta)/(g)`
C. `h=(2v^(2))/(g) sin theta (cos theta +a/g sin theta)`
D. `h=(2v^(2))/(a) sin theta (cos theta+g/a sin theta)`

Answer

D