Question

A particle is released from height ‘h’ above the surface of the earth. At certain height it’s K.E is 3 times of PE. The height from the surface of the earth and the speed of the Particle at the instant are respectively.

(1) \(\frac{h}{2}, \sqrt{\frac{3gh}{2}}\)

(2) \(\frac{h}{4}, \sqrt{\frac{3gh}{2}}\)

(3) \(\frac{h}{2}, \sqrt{\frac{3gh}{4}}\)

(4) \(\frac{h}{4} ,\sqrt{\frac{3gh}{4}}\)

Answer 1

Correct option is : (2) \(\frac{h}{4}, \sqrt{\frac{3gh}{2}}\) 

mgh = KE + PE

mgh = 4mgh'

\(h' = \frac{h}{4} \)

\(\frac{1}{2}mv^2 = 3 \times mg \frac{h}{4}\)

\(v = \sqrt{\frac{3}{2}gh}\)