Kinematics

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Question 201 of 308

Two projectiles are projected with the same initial velocities at the 15∘15^\circ and 30∘30^\circ with respect to the horizontal. The ratio of their ranges is 1:x1:x. The value of xx is:

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Answer: (B) 3\sqrt{3}

The horizontal range of a projectile is given by the formula R=u2sin⁡(2θ)gR = \dfrac{u^2 \sin(2\theta)}{g}.

For the first projectile, the angle of projection is θ1=15∘\theta_1 = 15^\circ.

R1=u2sin⁡(2×15∘)g=u2sin⁡(30∘)g=u22gR_1 = \dfrac{u^2 \sin(2 \times 15^\circ)}{g} = \dfrac{u^2 \sin(30^\circ)}{g} = \dfrac{u^2}{2g}

For the second projectile, the angle of projection is θ2=30∘\theta_2 = 30^\circ.

R2=u2sin⁡(2×30∘)g=u2sin⁡(60∘)g=3u22gR_2 = \dfrac{u^2 \sin(2 \times 30^\circ)}{g} = \dfrac{u^2 \sin(60^\circ)}{g} = \dfrac{\sqrt{3}u^2}{2g}

The ratio of their ranges is:

R1R2=u22g3u22g=13\dfrac{R_1}{R_2} = \dfrac{\dfrac{u^2}{2g}}{\dfrac{\sqrt{3}u^2}{2g}} = \dfrac{1}{\sqrt{3}}

Given that the ratio is 1:x1:x, we get x=3x = \sqrt{3}.

Answer: 3\sqrt{3}

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