A ball falls on the surface from 10 m height and rebounds to 2.5 m. If the duration of contact with the floor is 0.01 sec. Then average acceleration during contact is:

Options

A. 2100 `ms^(-2)`
B. 1400 `ms^(-2)`
C. 700 `ms^(-2)`
D. 400 `ms^(-2)`

Answer

a

Explanation

Here `a=(V_2-V_1)/(t)`
Now `v_2=sqrt(2gh_2)=sqrt(2xx10xx2.5)=sqrt(50)`
and `v_1=sqrt(2xx10xx10)=sqrt(200)`
`:. (sqrt(200)-sqrt(50))/(0.01)=a`
or a=2100 `ms^(-2)`