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(C) The potential energy $(PE)$ at maximum height $(H)$ is given by $PE = mgH$.For a projectile launched with velocity $u$ at an angle $	heta$ with t

Vedclass · Accepted Answer

$4 : 1$

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(C) The potential energy $(PE)$ at maximum height $(H)$ is given by $PE = mgH$.For a projectile launched with velocity $u$ at an angle $	heta$ with the horizontal,the maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$.Thus,$PE = mg \left( \frac{u^2 \sin^2 	heta}{2g} ight) = \frac{1}{2} m u^2 \sin^2 	heta$.For the first ball thrown vertically upward,the angle with the horizontal is $	heta_1 = 90^\circ$. Therefore,$(PE)_1 = \frac{1}{2} m u^2 \sin^2(90^\circ) = \frac{1}{2} m u^2$.For the second ball thrown at an angle of $60^\circ$ with the vertical,the angle with the horizontal is $	heta_2 = 90^\circ - 60^\circ = 30^\circ$. Therefore,$(PE)_2 = \frac{1}{2} m u^2 \sin^2(30^\circ) = \frac{1}{2} m u^2 (\frac{1}{2})^2 = \frac{1}{8} m u^2$.The ratio of their potential energies is $\frac{(PE)_1}{(PE)_2} = \frac{\frac{1}{2} m u^2}{\frac{1}{8} m u^2} = \frac{8}{2} = 4:1$.

Two balls are thrown with the same velocity,one vertically upward and the other at an angle of $60^\circ$ with the vertical. What is the ratio of their potential energies at their respective maximum heights?

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