A particle is released freely from a height H . At a certain height, its kinetic energy is two times of its potential energy. Then, the height and the speed of the particle at that instant are respectively ( g= acceleration due to gravity)

Step-by-step solution

If particle falls by a distance x , then aligned KE = 1 2 m v^2 & = 1 2 m(2 g x)=m g x PE & =m g(H-x) aligned As, KE =2( PE ) m g x=2 m g(H-x) x= 2 H 3 So, height of particle is H-x= H 3 . Speed of particle at H 3 distance = 2 g ( 2 H 3 ) =2 g H 3