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  4. Find angle of projection with the horizontal in terms of maximum height attained and horizontal range.

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Q. Find angle of projection with the horizontal in terms of maximum height attained and horizontal range.

Motion in a Plane

Solution:

Maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$ .....(i)
Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{g}$ .....(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{H}{R}=\frac{\tan \theta}{4}$
$\theta=\tan ^{-1} \frac{4 H}{R}$