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Motion in a Plane

Section: Physics

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Question : 50 of 94

Marks: +1, -0

Two projectiles

P_1

P_2

thrown with speed in the ratio

\sqrt{3}: \sqrt{2}

, attain the same height during their motion. If

P_2

is thrown at an angle of

60^{\circ}

with the horizontal, the angle of projection of

P_1

with horizontal will be :

[30-Jun-2022-Shift-1]

$15^{\circ}$
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$

Solution:

We know,

Maximum height of a projectile

(H)=\; \frac{u^{2} \sin^{2} \theta}{2 g}

Given, Ratio of initial velocity of two projectile

\; \frac{u_1}{u_2} = \; \frac{\sqrt{3}}{\sqrt{2}}

Both projectile reach the same maximum height.

\therefore H_1 = H_2

\Rightarrow \; \frac{u_1^2 \sin^2 \theta_1}{2 g} = \; \frac{u_2^2 \sin^2 \theta_2}{2 g}

\Rightarrow u_1^2 \sin^2 \theta_1 = u_2^2 \sin^2 \theta_2

\Rightarrow \left(\; \frac{u_1}{u_2}\right)^2 = \left(\; \frac{\sin \theta_2}{\sin \theta_1}\right)^2

\Rightarrow \left(\; \frac{\sqrt{3}}{\sqrt{2}}\right)^2 = \left(\; \frac{\sin 60^{\circ}}{\sin \theta_1}\right)^2

\Rightarrow \; \frac{3}{2} = \; \frac{3}{4 \sin^2 \theta_1}

\Rightarrow \sin_{\theta_1}^2 = \; \frac{1}{2}

\Rightarrow \sin \theta_1 = \; \frac{1}{\sqrt{2}} = \sin 45^{\circ}

\Rightarrow \theta_1 = 45^{\circ}

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