Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds thr other by an amount equal to half the sum of the height attained by them. Then angle of projection of the stone which attains smaller height is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `tan^(-1)(3//4)`

Answer

C

Explanation

It in given that, `H_(1) - H_(2) = (H_(1)+H_(2))/(2)` or `H_(1) = 3H_(2)`
`:. (u^(2)sin^(2)theta)/(2g) = 3 {(u^(2)sin^(2)(90^(@)-theta))/(2g)}`
`tan^(2) theta = 3`
`:. tan theta = sqrt(3)` or `theta = 60^(@)`
Therefore, the other angle is `90^(@) - theta` or `30^(@)`.
Therefore, the other angle is `90^(@) - theta` or `30^(@)`.