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Question

(C) Let the mass of each ball be $m$ and the projection speed be $u$.For the first ball projected vertically upwards, the maximum height reached is $H

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(C) Let the mass of each ball be $m$ and the projection speed be $u$.For the first ball projected vertically upwards, the maximum height reached is $H_1 = \frac{u^2}{2g}$. The potential energy at the highest point is $PE_1 = mgH_1 = mg \left(\frac{u^2}{2g}ight) = \frac{1}{2}mu^2$.For the second ball projected at an angle $	heta = 60^{\circ}$ with the vertical, the angle with the horizontal is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The maximum height reached is $H_2 = \frac{u^2 \sin^2 \alpha}{2g} = \frac{u^2 \sin^2 30^{\circ}}{2g} = \frac{u^2 (1/2)^2}{2g} = \frac{u^2}{8g}$.The potential energy at the highest point is $PE_2 = mgH_2 = mg \left(\frac{u^2}{8g}ight) = \frac{1}{8}mu^2$.The ratio of their potential energies is $\frac{PE_1}{PE_2} = \frac{\frac{1}{2}mu^2}{\frac{1}{8}mu^2} = \frac{8}{2} = \frac{4}{1}$.

Two balls of same mass are projected with the same speed, one vertically upwards and the other at an angle of $60^{\circ}$ with the vertical. The ratio of their potential energies at the highest point is:

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