A body is projected horizontally with a velocity of `u ms^(-1)` at an angle `theta` with the horizontal. The kinetic energy at the highest point is `(3)/(4)th` of the kinetic energy. The value of `theta` is :

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `120^(@)`

Answer

A

Explanation

`(KE_(1))/(KE_(2)) = 4/3 rArr (1/2 mu^(2))/(1/2 m(u cos beta)^(2)) = 4/3`
`rArr 3u^(2) =4u^(2) cos^(2)beta`
`rArr cos^(2)beta = 3/4 rArr cosbeta = sqrt(3/4)`
`rArr cos beta = sqrt(3)/2 rArr beta = 30^(@)`