Two balls are projected with the same velocit | vedclass

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(B) Since the ranges are equal for the same projection velocity,the angles of projection must be complementary.Let the angles be $	heta_1$ and $	het

Vedclass · Accepted Answer

$3h$

Vedclass · Answer

(B) Since the ranges are equal for the same projection velocity,the angles of projection must be complementary.Let the angles be $	heta_1$ and $	heta_2$. Then $	heta_1 + 	heta_2 = 90^{\circ}$.Given $	heta_1 = 30^{\circ}$,therefore $	heta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$.The formula for the maximum height of a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first ball,$h = \frac{u^2 \sin^2(30^{\circ})}{2g} = \frac{u^2}{2g} \left(\frac{1}{2}ight)^2 = \frac{u^2}{8g}$.For the second ball,$H_2 = \frac{u^2 \sin^2(60^{\circ})}{2g} = \frac{u^2}{2g} \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{u^2}{2g} \cdot \frac{3}{4} = \frac{3u^2}{8g}$.Comparing the two,$H_2 = 3 \left(\frac{u^2}{8g}ight) = 3h$.

Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $30^{\circ}$ and its maximum height is $h$,then the maximum height of the other will be

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