A body of mass 10 kg falls from a height of 10 m. What will be its kinetic energy just before it touches the ground ?
Prove that, this kinetic energy is equal to the potential stored at the initial position of the body. `[g=980 cm *s^(-2)`]

Explanation

Potential energy at the initial position = mgh `=10 xx 9.8 xx10 =980 J` .
Let the final velocity of the body just before it touches the ground be v .
So, `v^2 =u^2+2gh =0 +2xx9.8xx10=196 m^2*s^(-2)`
`therefore` Kinetic energy of the body just before it touches ground
`=1/2 mv^2 =1/2 xx10xx 196=980 J`
Hence, kinetic energy gained =initial potential energy.