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 -
- S4, 3gS2
- S4, 3gS2
- S4, 3gS2
- S2, 3gS2
Answer
(1) S4, 3gS2
Solution
Take potential energy zero at the surface of the Earth. Let the height be h when KE = 3 PE.
By conservation of mechanical energy:
mgS = KE + PE = 3mgh + mgh = 4mgh
⇒ h = S4
Speed after falling a distance S - h = 3S4:
v2 = 2g 3S4 = 3gS2
∴ v = 3gS2
Answer: (1)