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: (C)

1:07

Answer 2

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