Question

A body is thrown vertically up with certain initial velocity. The potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

• A. 1:01
• B. 1:04
• C. 1:07
• D. 1:08

Answer 1

Answer: (B)

1:04

Answer 2

In first case, let at point P, its kinetic and potential energies are equal ie, $\frac{1}{2}m{{v}^{2}}=mgh$ $\Rightarrow$ $h=\frac{{{v}^{2}}}{2g}$ ?(i) In second case, when bodys velocity is 2v then at the same point P $\frac{PE}{KE}=\frac{mg\times \frac{{{v}^{2}}}{2g}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{1}{4}$