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(B) If the range is the same for two projectiles thrown with the same speed,the angles of projection must be complementary,i.e.,$	heta$ and $(90^\cir

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$\frac{u^2}{2g}$

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(B) If the range is the same for two projectiles thrown with the same speed,the angles of projection must be complementary,i.e.,$	heta$ and $(90^\circ - 	heta)$.The maximum height attained by a projectile is given by the formula $h = \frac{u^2 \sin^2 	heta}{2g}$.For the first stone,$h_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second stone,$h_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Adding the two heights:$h_1 + h_2 = \frac{u^2 \sin^2 	heta}{2g} + \frac{u^2 \cos^2 	heta}{2g}$$h_1 + h_2 = \frac{u^2}{2g} (\sin^2 	heta + \cos^2 	heta)$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$h_1 + h_2 = \frac{u^2}{2g}$.

Two stones are thrown with the same speed $u$ at different angles from the ground into the air. If both stones have the same range and the heights attained by them are $h_1$ and $h_2$,then $h_1 + h_2$ is equal to .......

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