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(C) For two projectiles to have the same range with the same initial speed,the sum of their projection angles must be $90^{\circ}$.Given $	heta_1 = 6

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) For two projectiles to have the same range with the same initial speed,the sum of their projection angles must be $90^{\circ}$.Given $	heta_1 = 60^{\circ}$,therefore $	heta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.The maximum height attained by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.The sum of the maximum heights is $H_1 + H_2 = \frac{u^2 \sin^2 	heta_1}{2g} + \frac{u^2 \sin^2 	heta_2}{2g} = \frac{u^2}{2g} (\sin^2 60^{\circ} + \sin^2 30^{\circ})$.Substituting the values: $H_1 + H_2 = \frac{40^2}{2 	imes 10} ((\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2) = \frac{1600}{20} (\frac{3}{4} + \frac{1}{4}) = 80 	imes 1 = 80 \ m$.

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