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. `2100 m// sec^(2) ` downwards
B. ` 2100m//sec^(2) ` upwards
C. ` 1400m //sec ^(2) ` upwards
D. ` 700 m// sec ^(2) ` upwards

Answer

B

Explanation

Velocity at the time of striking the floor
` v= sqrt ( 2gh_1) =sqrt ( 2xx 9.8xx 10)= 14 m//s " "`Velocity with which it rebounds.
` therefore " " ` Avg. acceleration ` =(Delta v)/(Delta t) = ( 21)/( 0.01) =2100 m//s^(2) ` ( upwards )