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(A) The potential energy $PE$ at the highest point is given by $PE = mgh$,where $h$ is the maximum height.For ball $A$,projected vertically ($90^{\cir

Vedclass · Accepted Answer

$4: 3$

Vedclass · Answer

(A) The potential energy $PE$ at the highest point is given by $PE = mgh$,where $h$ is the maximum height.For ball $A$,projected vertically ($90^{\circ}$ to horizontal),the maximum height is $h_1 = \frac{u^2}{2g}$.For ball $B$,projected at an angle of $30^{\circ}$ with the vertical,the angle with the horizontal is $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.The maximum height for ball $B$ is $h_2 = \frac{u^2 \sin^2(60^{\circ})}{2g} = \frac{u^2}{2g} 	imes (\frac{\sqrt{3}}{2})^2 = \frac{3u^2}{8g}$.The ratio of potential energies is $\frac{PE_A}{PE_B} = \frac{mgh_1}{mgh_2} = \frac{h_1}{h_2}$.Substituting the values: $\frac{h_1}{h_2} = \frac{u^2}{2g} 	imes \frac{8g}{3u^2} = \frac{8}{6} = \frac{4}{3}$.Thus,the ratio is $4:3$.

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