Q. 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 highest points of their journey, the ratio of their potential energies will be
A) $1:1$
B) $2:1$
C) $3:2$
D) $4:1$
Explanation
(d) For first ball, $mgh_{1} = \frac{1}{2}mu^{2}$
[image:6_image61.emf]
$i.e.,$ $h_{1} = \frac{u^{2}}{2g}$
For second ball
$mgh_{2} = mg\frac{u^{2}\cos^{2}\theta}{2g} = \frac{1}{2}mu^{2}\cos^{2}\theta = \frac{1}{2}mu^{2}\cos^{2}{60{^\circ}}$
$= \frac{1}{2}mu^{2}\left( \frac{1}{2} \right)^{2} = \frac{1}{2}mu^{2}\left( \frac{1}{4} \right)$
[image:6_image62.emf]
$\Rightarrow h_{2} = \frac{u^{2}}{8g}$
$\therefore\frac{h_{1}}{h_{2}} = \frac{u^{2}}{2g} \times \frac{8g}{u^{2}} \Rightarrow \frac{h_{1}}{h_{2}} = \frac{4}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.