Motion in a Plane - Result Question 31
34. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is :
[2012]
(a) $\theta=\tan ^{-1}(\frac{1}{4})$
(b) $\theta=\tan ^{-1}$
(c) $\theta=\tan ^{-1}(2)$
(d) $\theta=45^{\circ}$
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Solution:
- (b) Horizontal range,
$R=\frac{u^{2} \sin 2 \theta}{g}$
Maximum height,
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
According to the condition,
$ R=H $
$\Rightarrow \frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow 2 \sin \theta \cos \theta=\frac{\sin ^{2} \theta}{2}$
$2 \cos \theta=\frac{\sin \theta}{2}$
$\Rightarrow \cot \theta=\frac{1}{4}$
$\Rightarrow \tan \theta=4$
$\Rightarrow \theta=[\tan ^{-1}(4)]$
In case of projective motion range $R$ is $n$ times the maximum height $H$
i.e., $R=n H \Rightarrow \frac{u^{2} \sin 2 \theta}{g}=n \frac{u^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow Tan \theta=\frac{4}{n}$ Angle of projection $\theta=\tan ^{-1}(\frac{4}{n})$
