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(C) Let the angles of projection be $	heta$ and $(90^{\circ}-	heta)$. The horizontal ranges are equal,which is consistent with these complementary a

Vedclass · Accepted Answer

$30^{\circ}$

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(C) Let the angles of projection be $	heta$ and $(90^{\circ}-	heta)$. The horizontal ranges are equal,which is consistent with these complementary angles.Let $H_1$ be the maximum height of the first stone and $H_2$ be the maximum height of the second stone,where $H_1 > H_2$.The given condition is: $H_1 - H_2 = \frac{1}{2}(H_1 + H_2)$.Multiplying by $2$,we get $2H_1 - 2H_2 = H_1 + H_2$,which simplifies to $H_1 = 3H_2$.The formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$.Substituting this into the equation: $\frac{u^2 \sin^2 	heta}{2g} = 3 \left( \frac{u^2 \sin^2(90^{\circ}-	heta)}{2g} ight)$.This simplifies to $\sin^2 	heta = 3 \cos^2 	heta$,or $	an^2 	heta = 3$.Thus,$	an 	heta = \sqrt{3}$,which gives $	heta = 60^{\circ}$.The two angles are $60^{\circ}$ and $30^{\circ}$. The stone with the smaller height corresponds to the smaller angle of projection,which is $30^{\circ}$.

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

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