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Q.
The angle for which maximum height and horizontal range are same for a projectile is
BHUBHU 2007
Solution:
Vertical component of velocity at the highest point is zero.
Let body is projected with an initial velocity $u$ making an angle $\theta$ with the horizontal.
The vertical velocity at $O$ is zero.
From $v^{2}=u^{2}-2 g h$
$v=v_{y}=0$ and $u=u_{y}=u \sin \theta$
$0=(u \sin \theta)^{2}-2 g H$
$\Rightarrow H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
Also, range $=$ horizontal velocity $\times$ time of flight
$R=u_{x} \times T=(u \cos \theta) \times \frac{2 u \sin 2 \theta}{g}$
$R=\frac{u^{2} \sin 2 \theta}{g}$
Given, $H=R$
$\therefore \frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2} \sin 2 \theta}{g}$
$\frac{\sin ^{2} \theta}{2}=2 \sin \theta \cos \theta$
$\frac{\sin ^{2} \theta}{\cos \theta}=4$
$\Rightarrow \tan \theta=4$
$\therefore \theta=\tan ^{-1}(4), \theta \approx 76^{\circ}$
