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

Explanation

A particle is …………
Since x - displecement is zero
`0 = Vsin theta t-(1)/(2)at^(2)`
`t = (2Vsin theta)/(a)`
`h = Vcos theta t+(1)/(2)"gt"^(2)`
`h = (2V^(2))/(a)sin theta [cos theta+(g)/(a)sin theta]`