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(C) The formula for the maximum height $H$ attained by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Since both projectiles are projecte

Vedclass · Accepted Answer

$1: 3$

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(C) The formula for the maximum height $H$ attained by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Since both projectiles are projected with the same speed $u$,the ratio of their maximum heights $H_1$ and $H_2$ is given by $\frac{H_1}{H_2} = \frac{\sin^2 	heta_1}{\sin^2 	heta_2}$.Given $	heta_1 = 30^{\circ}$ and $	heta_2 = 60^{\circ}$,we have $\sin 30^{\circ} = \frac{1}{2}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Substituting these values,we get $\frac{H_1}{H_2} = \frac{(\sin 30^{\circ})^2}{(\sin 60^{\circ})^2} = \frac{(1/2)^2}{(\sqrt{3}/2)^2} = \frac{1/4}{3/4} = \frac{1}{3}$.Thus,the ratio is $1:3$.

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

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