JEE Main 2025 · 3 April · morning · Single correct · +4 / −1
A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.
- (a) S/4, (3 gS)/2
- (b) S/2, (3 gS)/2
- (c) S/4, √((3 gS)/2)
- (d) S/2, √((3 gS)/2)
Answer and solution
Answer (c) S/4, √((3 gS)/2)
Solution
Physics sir · 4 steps
Total energy stays mgS all the way down, so split it in the ratio 3 : 1 between KE and PE.
- Total energy. The particle starts at rest at height S, so its total mechanical energy is mgS. With no air resistance, this total stays the same throughout the fall.
- Find the height. At height h, PE = mgh and KE = 3 PE = 3mgh. So the total is mgh + 3mgh = 4mgh = mgS, which gives h = S/4.
1/2mv² = 3mg(S/4) = 3mgS/4
- Find the speed. Cancel m and multiply both sides by 2 to get v² = 3gS/2. So v = √(3gS/2).
- So. The particle is at height S/4 and moving at √(3gS/2), which is option (c).
The idea to keep. If KE is n times PE, the total energy splits into n + 1 equal parts. PE takes one part, so the height is the starting height divided by n + 1.
Answer from the NTA final answer key. Solution by PrepWiser.