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:

  1. (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 $\theta=\tan ^{-1}(\frac{4}{n})$