Question

On the basis of conservation of energy prove that velocity of a body on striking the ground from height h is
$\sqrt{2gh}$

Answer 1

Let the potential energy at v = 0 be, $P = mgh$

Let the kinetic energy at the time of striking be, $K = 1/2mv^{2}$

By law of conservation of energy,

$P = K\\mgh = 1/2mv^{2}\\gh = 1/2v^{2}\\v^{2} = 2gh\\v = \sqrt{2gh}$

Thus by Law of Conservation of Energy,  velocity of a body on striking the ground from height h is $\sqrt{2gh}$ 

Answer 2

Assuming that you threw a stone upwards, and velocity of the stone becomes v =0 at a height h, then falls on the ground to attain speed u,

u² = v² - 2gh

u = √2gh                                   .......................... since v = 0