In a projectile motion, the maximum height reached equals to the horizontal range covered by the body, the angle of projection with the horizontal equals to

Options

A. `tan^-1 1`
B. `tan^-1` 1/4
C. `tan^-1` 2
D. `tan^-1`(4)

Answer

D

Explanation

maximum height =`(u^2 sin^2 theta)/(2g)`
Horizontal Range = `(u^2 sin2 theta)/(g)`
Given `(u^2 sin^2 theta)/(2g)` = `(u^2 sin2 theta)/(g)` or `sintheta = 2 sin 2 theta` or `tantheta` = 4 or `theta tan^-1(4)`.