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Q.
The angle for which maximum height and horizontal range are same for a projectile is:
AFMCAFMC 2001
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 \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 \theta}{\cos \theta} =4$
$ \tan \theta =4 $
$ \Rightarrow \theta =\tan ^{-1}(4) \theta \approx 76^{\circ} $
