Question

A particle is released from a height h. At a certain height,
its KE is two times its potential energy. Height and
speed of the particle at that instant are
​

Answer 1

Answer:

me a few years old school for a couple

Answer 2

Explanation:

Given,

Height = h

A/q,

2 K.E. = mgh

=> 2*1/2 mv^2 = mgh

=>v^2 =gh

=> v =

$\sqrt{gh}$

and, h = v^2/g