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

J & K CETJ & K CET 2004Motion in a Plane

Solution:

Let u be initial velocity of projection at angle $\theta$ with the horizontal.
Then, horizontal range, $R=4 \sqrt{3}$
$\therefore \frac{u^{2} \sin 2 \theta}{g}=4 \sqrt{3} \cdot \frac{u^{2} \sin ^{2} \theta}{2 g}$
or $2 \sin \theta \cos \theta=2 \sqrt{3} \sin ^{2} \theta$
or $\frac{\cos \theta}{\sin \theta}=\sqrt{3}$
$\cot \theta=\sqrt{3}$
$\theta =30^{\circ}$.

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