Question

The angle for which maximum height and horizontal range are same for a projectile is

• A. $32{}^{\circ }$
• B. $48{}^{\circ }$
• C. $76{}^{\circ }$
• D. $84{}^{\circ }$

Answer 1

Answer: (C)

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 $v^2=u^2-2gh$
$v=v_y=0$ and $u=u_y=u\sin \theta$
$0=(u\sin \theta {)}^2-2gH$
$\Rightarrow H=\frac{u^2{\sin }^2\theta }{2g}$
Also, range $=$ horizontal velocity $\times$ time of flight
$R=u_x\times T=(u\cos \theta )\times \frac{2u\sin 2\theta }{g}$
$R=\frac{u^2\sin 2\theta }{g}$
Given, $H=R$
$\therefore \frac{u^2{\sin }^2\theta }{2g}=\frac{u^2\sin 2\theta }{g}$
$\frac{{\sin }^2\theta }{2}=2\sin \theta \cos \theta$
$\frac{{\sin }^2\theta }{\cos \theta }=4$
$\Rightarrow \tan \theta =4$
$\therefore \theta ={\tan }^{-1}(4),\theta \approx 76^{\circ }$