Question

A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of particle at that instant are respectively.​

Answer 1

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Height S, its PE = m g S
Let x be the distance through which particle has travelled from the start
Then potential energy = m g (S - x)
By conservation of energy mg S - m g ( S-x) = m g x will be the KE
Given KE = 3 * PE 
So m g x = 3 m g ( S - x)
OR x = 3 S - 3 x
==> 4 x = 3 S
So x = 3/4 * S
Hence the height is S - 3/4 S = 1/4 * S 
Speed = v^2 = 2 g *3/4 S 
So v = (3/2 * g S)^1/2

Answer 2

Height S, its PE = m g S

Let x be the distance through which particle has travelled from the start

Then potential energy = m g (S - x)

By conservation of energy mg S - m g ( S-x) = m g x will be the KE

Given KE = 3 * PE

So m g x = 3 m g ( S - x)

OR x = 3 S - 3 x

==> 4 x = 3 S

So x = 3/4 * S

Hence the height is S - 3/4 S = 1/4 * S

Speed = v^2 = 2 g *3/4 S

So v = (3/2 * g S)^1/2