Two stones are projected so as to reach same distance from point of projection on a horizontal surfa

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3-2.Motion in Plane

hard

Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the height attained by them. Then, angle of projection of the stone which attains smaller height is $........$

A

$45$

B

$60$

C

$30$

D

$\tan ^{-1} (\frac{3}{4})$

Solution

(c)

$H_1-H_2=\frac{H_1+H_2}{2}$ or $H_1=3 H_2$

$\therefore \quad \frac{u^2 \sin ^2 \theta}{2 g}=2\left\{\frac{u^2 \sin ^2\left(90^{\circ}-\theta\right)}{2 g}\right\}$

$\tan ^2 \theta=3$

$\therefore \tan \theta =\sqrt{3}$

$\text { or } \theta =60^{\circ}$

Therefore, the other angle is $\left(90^{\circ}-\theta\right)$ or $30^{\circ}$

Standard 11

Physics

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