The horizontal range of a projectile is 4 √3 times of its maximum height. The angle of projection will be

Correct answer: (C) 30^°. R=4 √3 (2 u^2 sin θ cos θ)/g =4 √3 ((u^2 sin^2 θ)/(2 g) ) tan θ =1/(√3) ∴ θ =30^°

Q81 / 110 · VITEEE 2024 · Physics

The horizontal range of a projectile is $4 \sqrt{3}$ times of its maximum height. The angle of projection will be

A
$60^{\circ}$
B
$37^{\circ}$
Correct answer
C
$30^{\circ}$
D
$45^{\circ}$

Solution

$R=4 \sqrt{3}$ $\begin{aligned} \frac{2 u^2 \sin \theta \cos \theta}{g} & =4 \sqrt{3}\left(\frac{u^2 \sin ^2 \theta}{2 g}\right) \\ \tan \theta & =\frac{1}{\sqrt{3}} \\ \therefore \quad \theta & =30^{\circ}\end{aligned}$

Hence, the correct answer is option: 'C'.

The horizontal range of a projectile is $4 \sqrt{3}$ times of its maximum height. The angle of projection will be

Solution

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