Form a height of 20 m above a horizontal floor , a ball is thrown down with initial velocity 20 m /s . After striking the floor , the ball bounces to the same height from which it was thrown . Find the coefficient of restitution for the collision between the ball and the floor ? (g= 10 `m//s^(2)`)

Explanation

Initial velocity `u^(1)=20m//s,h=20mg=10m//s^(2)`
Velocity of approach ,
`u^(1)=u^(1^(2))=u^(1^(2))+2as=400+2xx10xx20`
`rArru^(2)=400+400=800rArru=20sqrt(2)`
Height of rebounce = h = 20m.
`therefore` Velocity of separation
`v=sqrt(2gh)=sqrt(2xx10xx20)=sqrt(400)=20`
Coefficient of restitution.
`e=("Velocity of separation")/("Velocity of approach")=(20)/(20sqrt(2))=(1)/(sqrt(2))`