A body is dropped from 1000m height,if its potential energy is 8*10power8 . What will be its velocity on reaching the ground.?

Take g = 10 m/s²

Potential energy:
m g h = 8 × 10⁸ J
m × 10 × 1000 = 8 × 10⁸ J
Mass, m = 80000 kg

K.E. on reaching the ground = Loss in P.E.
(1/2) m v² = 8 × 10⁸ J
(1/2) × 80000 × v² = 8 × 10⁸
v² = 2 × (8 × 10⁸) / 80000
v = √[2 × (8 × 10⁸) / 80000]
Velocity on reaching the ground, v = 141 m/s

On reaching the ground, it's velocity will cease to exist..

Conservation of energy applies. Initial potential and kinetic energy equal final potential and kinetic energy.
Potential energy = mgh
Kinetic energy = (mv²) / 2

The body was dropped and starts with 0 velocity, so there is no initial kinetic energy.
Likewise when the body hits the ground, there is no longer any potential energy.

g = 9.81 m/s²

Truth be told, the given initial potential energy is not needed since we know the acceleration due to gravity and the height already.

mgh = (mv²) / 2
gh = v² / 2
(9.81)(1000) = v² / 2
19620 = v²
v = 140 m/s