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(D) For two projectiles with the same initial speed $u$ and equal ranges,the angles of projection $	heta_1$ and $	heta_2$ must be complementary,i.e.

Vedclass · Accepted Answer

$h_1/3$

Vedclass · Answer

(D) For two projectiles with the same initial speed $u$ and equal ranges,the angles of projection $	heta_1$ and $	heta_2$ must be complementary,i.e.,$	heta_1 + 	heta_2 = \pi / 2$.Given $	heta_1 = \pi / 3$,then $	heta_2 = \pi / 2 - \pi / 3 = \pi / 6$.The formula for maximum height is $h = \frac{u^2 \sin^2 	heta}{2g}$.Therefore,the ratio of the heights is $\frac{h_2}{h_1} = \frac{u^2 \sin^2 	heta_2 / 2g}{u^2 \sin^2 	heta_1 / 2g} = \frac{\sin^2 	heta_2}{\sin^2 	heta_1}$.Substituting the values: $\frac{h_2}{h_1} = \frac{\sin^2(\pi / 6)}{\sin^2(\pi / 3)} = \frac{(1/2)^2}{(\sqrt{3}/2)^2} = \frac{1/4}{3/4} = \frac{1}{3}$.Thus,$h_2 = h_1 / 3$.

Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $\pi / 3$ and its maximum height is $h_1$,then the maximum height of the other will be:

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