Question

When a body is projected vertically up from the ground with certain velocity, its potential energy and kinetic energy at a point A are in the ratio $2:3$. If the same body is projected with double the previous velocity, then at the same point A the ratio of its potential energy to kinetic energy is

• A. 9 : 1
• B. 2 : 9
• C. 1 : 9
• D. 9 : 2
• E. 3 : 2

Answer 1

Answer: (C)

Let in the first case, the total energy at the throwing point be $5E$.
$KE$ at a particular height $=3E$, potential energy at the height $=2E$
As velocity of throw is doubled, kinetic energy becomes 4 times.
$\therefore$ Total $KE$ at the throwing point $20E$
$PE$ at the same height $=2E$ (will remain same as the previous value) and $KE$ at the same height $=18E$
$\frac{PE}{KE}=\frac{2E}{18E}=\frac{1}{9}=1:9$