A stone is projected vertically upwards with speed ' \(v\) '. Another stone of same mass is projected at an angle of \(60^{\circ}\) with the vertical with the same speed ' \(v\) '. The ratio of their potential energies at the highest points of their journey is \(\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right.\), \(\left.\cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]\)

P.E. of the stone projected vertically is,
P.E. = mgh
But \(\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}\)
\(\begin{aligned}
\therefore \quad \text { P. } \mathrm{E}_1 & =\mathrm{mg}\left(\frac{\mathrm{v}^2}{2 \mathrm{~g}}\right) \\
& =\frac{\mathrm{mv}^2}{2}... (i)
\end{aligned}\)
For the second stone thrown at an angle \(\theta\) to the horizontal,
\(\begin{array}{ll}
& \mathrm{h}=\frac{\mathrm{v}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{v}^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}=\frac{\mathrm{v}^2}{8 \mathrm{~g}} \\
\therefore \quad & \text { P.E. } 2=\mathrm{mg}\left(\frac{\mathrm{v}^2}{8 \mathrm{~g}}\right)=\frac{\mathrm{mv}^2}{8}... (ii)
\end{array}\)
Dividing equation (i) by equation (ii) -
\(-\frac{P \cdot E_1}{P \cdot E_2}=\frac{\left(\frac{m v^2}{2}\right)}{\left(\frac{m v^2}{8}\right)}=4: 1\)