Q. The angle of projection for which the horizontal range is equal to the maximum height of a projectile is :
J & K CETJ & K CET 2000
Solution:
Let a particle be projected with initial velocity u at an angle $\theta$ with the horizontal.
Then, range $(R)$ of projectile is
$R=\frac{u^{2} \sin 2 \theta}{g} \ldots$ (i)
Maximum height $H=\frac{u^{2} \sin ^{2} \theta}{2 g} \ldots$ (ii)
Given, $R=H$,
$\therefore \frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow 2 \sin \theta \cos \theta=\frac{\sin ^{2} \theta}{2}$
$\Rightarrow 4 \cos \theta=\sin \theta$
$\Rightarrow \tan \theta=4$
$\Rightarrow \theta=\tan ^{-1}(4)$
