A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5 m if the ball is in contact with floor for 0.01 s then the average acceleration during contact is nearly

Options

A. `1400 m s^(-2)`
B. `2100 m s^(-2)`
C. `700 m s^(-2)`
D. `2800 m s^(-2)`

Answer

b

Explanation

`h=10m`
`v^(2)=0+2g(10)`
`v^(2)=196m//s`
v=14m/s (downwards)

`0=u^(2)-2g(2.5)`
`u^(2)=5g`
`u=sqrt(5g)j=7m//s` (upwards)
During the collision of ball with the ground in time
t=0.01s
`a=(dv)/(dt)=(u-(-v))/(t)-(7+14)/(0.01)=2100m//s^(2)`