Let a ball of mass m being dropped from a cliff of height H as shown in figure .
The total mechanical energies `E_(0),E_(h) and E_(H)` of the ball at the indicated heights , zero (ground level ) h and H respectively are .
Total mechanical energy at height H ,
`E_(H) =mgH +1/2 mv^(2)`
but a maximum height v = 0
Total energy at h height ,
`E_(h)` = potential energy + kinetic energy
` = mgh+1/2 mv_(h)^(2) ( :. "velocity of ball at height h = "v_(h))`
Total energy at ground ,
`E_(0) =/2 mv_(f)^(2)`
where `v_(f)` is the final velocity of ball when it hit the ground .
From the conservation of mechanical energy at height H ,
`E_(H) =E_(0)`
` :. mgH = 1/2 mv_(f)^(2)`
where `v_(f)` is the final velocity of ball when it hit the ground .
From the conservation of mechanical energy at height H ,
`E_(H) =E_(0)`
` :. mgH = 1/2 mv_(f)^(2)`
` :. v_(f) = sqrt(2gH)`
From the conservation of mechanical energy at H and h .
`E_(H) =E_(h)`
` :. mgH =mgh + 1/2 mv_(f)^(2)`
` :. mgH =mgh +1/2mv_(f)^(2)`
` :. gH -gh =1/2 v_(f)^(2)`
` :. 2g(H-h)=v_(f)^(2)`
` :. v_(f)=sqrt(2g(H-h))`
At the height H , the energy is purely potential energy at height h and is fully kinetic at ground level .
Hence , this illustrates the conservation of mechanical energy for a free fall body .
