A stone is projected up with a vertical velocity u, reaches upto a maximum height h. When it is at a height of 3h/4 from the ground, the ratio of KE and PE at that point is: (consider PE=0 at the point of projection)

Question

A stone is projected up with a vertical velocity u, reaches upto a maximum height h. When it is at a height of 3h/4 from the ground, the ratio of KE and PE at that point is: (consider PE=0 at the point of projection)
a]1:1 b]1:2 c]1:3 d]3:1

Henry · Answer

Let u = Vo = 10 m/s.
g = 10 m/s^2.

hmax=(V^2-Vo^2)/2g = (0-100)/-20 = 5 m.

(3/4)hmax = (3/4)*5 = 3.75 m.

V^2 = Vo^2 + 2g*h
V^2 = 100 - 20*3.75 = 25
V = 5 m/s. @ (3/4)hmax.

When h = 0,
KE + PE = 0.5M*V^2 + 0 = 0.5M*V^2 =
0.5M*10^2 = 50M. = KE.
M = Mass.

When h = (3/4)hmax,
KE + PE = 50M
0.5M*5^2 + PE = 50M
PE == 50M - 12.5M = 37.5M

KE/PE = 12.5M/37.5M = 1/3 or 1:3.