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The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
A
$\theta = 45^\circ$
B
$\theta = \tan^{-1}\left(\frac{1}{4}\right)$
C
$\theta = \tan^{-1}(4)$
D
$\theta = \tan^{-1}(2)$
Detailed Solution
Maximum height $H = \frac{u^2\sin^2\theta}{2g}$; range $R = \frac{2u^2\sin\theta\cos\theta}{g}$
H = R: $\frac{u^2\sin^2\theta}{2g} = \frac{2u^2\sin\theta\cos\theta}{g}$
$\sin\theta = 4\cos\theta \Rightarrow \tan\theta = 4$
$\theta = \tan^{-1}(4)$
H = R: $\frac{u^2\sin^2\theta}{2g} = \frac{2u^2\sin\theta\cos\theta}{g}$
$\sin\theta = 4\cos\theta \Rightarrow \tan\theta = 4$
$\theta = \tan^{-1}(4)$