Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$45^{\circ}$

$60^{\circ}$

$90^{\circ}$

$30^{\circ}$

Formulas: Maximum Height ( $H$ ) is given by $H = \frac{v_0^2 \sin^2 \theta}{2g}$ . Horizontal Range ( $R$ ) is given by $R = \frac{v_0^2 \sin 2\theta}{g} = \frac{2v_0^2 \sin \theta \cos \theta}{g}$ .
Given Condition: $R = 4\sqrt{3} H$ .
Substitution: $\frac{2v_0^2 \sin \theta \cos \theta}{g} = 4\sqrt{3} \left( \frac{v_0^2 \sin^2 \theta}{2g} \right)$
Simplification: Cancel common terms ( $v_0^2$ , $g$ , and one $\sin \theta$ ): $2 \cos \theta = 4\sqrt{3} \left( \frac{\sin \theta}{2} \right)$ $2 \cos \theta = 2\sqrt{3} \sin \theta$ $\frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}}$ $\tan \theta = \frac{1}{\sqrt{3}}$
Calculate Angle: $\theta = 30^{\circ}$ .

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