A ball P is projected at an angle of 60^{circ | vedclass

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(A) The potential energy at the highest point is given by $PE = mgh$, where $h$ is the maximum height reached.For ball $P$, the angle with the vertica

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(A) The potential energy at the highest point is given by $PE = mgh$, where $h$ is the maximum height reached.For ball $P$, the angle with the vertical is $60^{\circ}$, so the angle with the horizontal is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.The maximum height for ball $P$ is $H_P = \frac{u^2 \sin^2(30^{\circ})}{2g} = \frac{u^2 (0.5)^2}{2g} = \frac{u^2}{8g}$.For ball $Q$, which is projected vertically upwards, the angle with the horizontal is $90^{\circ}$.The maximum height for ball $Q$ is $H_Q = \frac{u^2 \sin^2(90^{\circ})}{2g} = \frac{u^2}{2g}$.The ratio of potential energy of ball $P$ to ball $Q$ is $\frac{PE_P}{PE_Q} = \frac{mgH_P}{mgH_Q} = \frac{H_P}{H_Q} = \frac{u^2/8g}{u^2/2g} = \frac{2}{8} = \frac{1}{4}$.

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