Question

A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of ${60}^{\circ }$ with the vertical with the same initial speed. At the highest point, the ratio of their potential energies will be

• A. 4:1
• B. 3:2
• C. 2:3
• D. 2:1

Answer 1

The correct option is A 4:1

The maximum height attained by the first ball is $h_1=\frac{u^2}{2g}$ where u is the initial speed of projection. The maximum height attained by the second ball is $(\therefore \theta ={90}^{\circ }-{60}^{\circ }={30}^{\circ })$
$h_2=\frac{u^{2}si{n}^2\left({30}^{\circ }\right)}{2g}=\frac{u^2}{8g}$
Now. PE. of ball 1 at height $h_1=mg{h}_1$ and that of ball 2 at height $h_2=-mg{h}_2.$ Therefore, the ratio of potential energies $=\frac{h_1}{h_2}=\frac{u^2}{2g}X\frac{8g}{u^2}=4$