2021 GravitationEnergy Conservation under Gravity Medium Question 18 of 66 in Gravitation

A particle is released from height $S$ from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively:

A $\frac{S}{4},\ \frac{\sqrt{3gS}}{2}$
B $\frac{S}{2},\ \frac{\sqrt{3gS}}{2}$
C $\frac{S}{4},\ \sqrt{\frac{3gS}{2}}$
D $\frac{S}{4},\ \frac{3gS}{2}$

The answer is C.

Solution

Total energy E = mgS. When KE = 3U: U + KE = 4U = E $4mgh = mgS \Rightarrow h = \frac{S}{4}$ The particle has fallen $\frac{3S}{4}$, so $v = \sqrt{2g\left(\frac{3S}{4}\right)} = \sqrt{\frac{3gS}{2}}$

Concept KE and PE during free fall

Gravitation in past papers

66 questions from this chapter have appeared across 26 exam years.