First slide

Projection Under uniform Acceleration

Question

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

Easy

A

45°
B

60°
C

90°
D

30°

Solution

$\begin{aligned} H = V^{2} \sin^{2} θ / 2 g \\ R = V^{2} \sin 2 θ / g \end{aligned} \frac{H}{R} = \frac{1}{4} \tan θ \frac{H}{4 \sqrt{3} H} = \frac{1}{4} \tan θ θ = 30^{\circ}$

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