1.

Tennis ball is dropped on the floor from a height of 100 m. It rebounds to a height of 100 m. Ir the ball was in contact with the floor for 3.16 3. then the average acceleration during the contact is:

Answer»

700 `m//s^(2)`
140 `m//s^(2)`
280`m//s^(2)`
28 `m//s^(2)`

SOLUTION :`v=SQRT(2gh) implies v=sqrt(2xx9.8xx100)=sqrt(1960)`.
After rebound, v=`-sqrt(1960)`

acceleration`=("CHANGE in velcocity")/("time")`
`(2sqrt(1960))/(3.16)=28 ms^(-2)`


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