question_answer

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) \[1:1\]                                  

B) \[1:4\]

C) \[1:7\]                                  

D) \[1:8\]

Correct Answer: B

Solution :

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\[P\]                 \[\frac{PE}{KE}=\frac{mg\times \frac{{{v}^{2}}}{2g}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{1}{4}\]