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(D) The first ball is projected vertically upwards,so its maximum height is given by $h = \frac{u^2}{2g}$.The second ball is projected at an angle $

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$4:1$

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(D) The first ball is projected vertically upwards,so its maximum height is given by $h = \frac{u^2}{2g}$.The second ball is projected at an angle $	heta = 30^{\circ}$ with the horizontal,so its maximum height is given by $h' = \frac{u^2 \sin^2 	heta}{2g} = \frac{u^2 \sin^2 30^{\circ}}{2g} = \frac{u^2}{2g} 	imes (\frac{1}{2})^2 = \frac{1}{4} 	imes \frac{u^2}{2g} = \frac{h}{4}$.The potential energy at the highest point is $PE = mgh$.The ratio of their potential energies is $\frac{PE_1}{PE_2} = \frac{mgh}{mgh'} = \frac{h}{h/4} = \frac{4}{1}$.

$A$ ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected with the same speed at an angle of $30^{\circ}$ with the horizontal. At the highest point,the ratio of their potential energies is

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