a ball is dropped on the floor fromthe height of a 10m .it rebounds to aheight of 2.5m .if the ball

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Use 3 eqn of motion for downward motion of ball

V1² = U² +2*a*s

V1² = 0 + 2*9.8*10

V1² = 196

V1 = 14 m/s.

For upward motion also use same eqn.

0 = U² - 2*9.8*2.5

U²=49

U = 7

Chenge in velocity during the contact =14 -(-7) =21m/s

also time of contact is .01 s.

but acceleration = change in velocity/ time taken.

therefore acceleration during contact is = 21/0.01 =2100 m/s²

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