Question

A ball of mass m is dropped from a height H. At height H/3, the ratio of its PE to KE?

Answer 1

PE=mg×H/3=mgH/3
KE=1/2mv^2
=1/2m(u^2+2g(H-H/3)){usind 3rd equation of motion}
=1/2m(0+2g×2H/3)
=2mgH/3
PE:KE=1:2
DoNe

Answer 2

Given:

Mass of the ball = m

Height from which it is dropped = H

To Find:

Ratio of PE to KE at height H/3

Solution:

The potential energy at H/3 will be =  Mg4/3

Kinetic energy = Ke = 1/2mv²

Te = Pe + Ke

MgH = MgH/3 + Ke

Ke = MgH - MgH/3

Ke = 2MgH/3

Now,

Pe:Ke =  MgH/3 : 2MgH/3

= 1:2

Answer: The ratio of PE: KE at H/3 is 1:2