Question

A ball is thrown vertically down from height of 80 m from the ground with an initial velocity 'v'. The ball hits the ground, loses `1/6`th of its total mechanical energy, and rebounds back to the same height. If the acceleration due to gravity is 10 ms^-2, the value of 'v' is

Options

• 12 ms^-1

• 18 ms^-1

• 25 ms^-1

• 30 ms^{- 1}

Answer 1

18 ms^-1

Explanation:

Total mechanical energy of ball,

T= `1/2` mv²= mgh

Total energy after the collision,

`5/6(1/2"mv" ^2+"mgh")`

The ball rebounds back to the same height after collision,

∴ `5/6(1/2m"v" ^2+"mgh")`=mgh

∴ `5/6(1/2"v" ^2+gh)`=gh,

`15/12"v" ^2 +5/6"gh"` =gh,

∴ `"5v" ^2/12="gh"- "5gh"/6`

∴ `"v" ^2="gh"/6 xx 12/5 = 2/5 "gh"`

∴ `"v"= sqrt("2gh"/5)`

= `sqrt("2×10×80"/5)`

= `8 sqrt (5)` m/s= 18m/s.