Two balls of same mass are projected one vertically upwards and the other at angle 60° with the vertical. The ratio of their potential energy at the highest point is

The correct option is 4 : 1.

Let the mass of both balls be $m$ and their initial projection speed be $u$.

For the first ball projected vertically upwards:
The angle of projection with the horizontal is ${\theta }_1=90°$.
The maximum height reached by a vertically projected body is given by the formula:
$h_1=\frac{u^2}{2g}$

For the second ball projected at an angle of $60°$ with the vertical:
The angle of projection with the horizontal is:
${\theta }_2=90°-60°=30°$.
The maximum height reached by a projectile is given by:
$h_2=\frac{u^2{\sin }^2{\theta }_2}{2g}$
Substituting ${\theta }_2=30°$ and $\sin 30°=\frac{1}{2}$:
$h_2=\frac{u^2}{2g}{\left(\frac{1}{2}\right)}^2=\frac{u^2}{8g}$

The potential energy ($U$) of a body of mass $m$ at height $h$ is given by $U=mgh$.
Since the masses of both balls are the same, the ratio of their potential energies at the highest points is equal to the ratio of their maximum heights:
$\frac{U_1}{U_2}=\frac{mg{h}_1}{mg{h}_2}=\frac{h_1}{h_2}$

Substituting the values of $h_1$ and $h_2$:
$\frac{U_1}{U_2}=\frac{\left(\frac{u^2}{2g}\right)}{\left(\frac{u^2}{8g}\right)}=\frac{8}{2}=4$

Thus, the ratio of their potential energy at the highest point is 4 : 1.