Question

From an elevated point A, a stone is projected vertically upward. When the stone reaches a distance h below A, its speed is triple of what it was at a height h above A. Show that the highest point reached by the stone above A is 5h/4.

Answer 1

Let's say speed at point A is V_A so, E_A = 1/2 MV^2_A E_B = mgh + 1/2 MV^2_B E_c = -mgh + 1/2 MV^2_c E_A = E_B = E_C (conservation of energy) so mgh + 1/2 MV^2_B = -mgh + 1/2 MV^2_C 4gh + V^2_B = V^2_C V_C = 3 V_B so 4gh = 8 V^2_B \r\n V_B = squareroot gh/2 so E_B = mgh + 1/2 M (squareroot gh/2)^2 E_B = mgh + mgh/4 E_B = 5 mgh/4 clearly this energy can be attained at maximum height at P E so mgH_max = 5 mgh/4 H_max = 5h/4