A body is thrown from the surface of the earth with velocity `(gR_(e))//12`, where `R_(e)` is the radius of the earth at some angle from the vertical. If the maximum height reached by the body is `R_(e)//4`, then the angle of projection with the vertical is

Options

A. `sin^(-1)((sqrt(5))/4)`
B. `cos^(-1)((sqrt(5))/4)`
C. `sin^(-1)((sqrt(3))/2)`
D. none of these

Answer

A

Explanation

Let the speed at the maximum height be `v_(1)` then from energy be coservation
`(mv^(2))/2-(GM_(m))/(R_(e))=(mv_(1)^(2))/2-(GMm)/(5R_(e)//4)`
From angular momentum conservation
`mvR_(e)sintheta=mv_(1)xx(5R_(e))/4`
Using `v=sqrt((gR_(e))/2)` and solving the equation, we get `theta=sin^(-1)((sqrt(5))/4),`