First slide

Projectile motion

Question

The angle of projection at which the horizontal range and the maximum height of a projectile are equal is

Moderate

A

$30^{0}$
B

$45^{0}$
C

$60^{0}$
D

$76^{0}$

Solution

$H = R$ {given}
$\Rightarrow$ $\frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2} \sin 2 θ}{g}$

$\Rightarrow$ $\sin^{2} θ = 2 (2 \sin θ \cos θ)$ ${\sin θ \neq 0}$

$\Rightarrow$ $\sin θ = 4 \cos θ$

$\Rightarrow$ $\tan θ = 4$

$\Rightarrow$ $∵ θ = \tan^{- 1} 4 = {75.96}^{0} \approx 76^{0}$

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