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A small ball is dropped from a height $100 m$ on a floor. The ball rebounds to a height of 20 metre. Calculate the average acceleration during the contact if the ball was in contact of the floor for $0.02 sec$, take $g=10 m / sec ^{2}$.

Solution —

The velocitywith which the ball strikes the ground.

$u=\sqrt{2 g h}=\sqrt{2 \times 10 \times 100}=44.72 m / sec \text {. }$

The velocity of its rebound

$v=\sqrt{2 g h_{1}}=\sqrt{2 \times 10 \times 20}=20 m / sec .$

The change in velocity $=v-(-u)=v+u$

$=20+44.72=64.72 m / sec \text {. }$

$\therefore \quad$ Therefore, average acceleration $=\frac{\text { change in velocity }}{\text { time }}=\frac{64 \cdot 72}{0 \cdot 02}$ $=3236 m / sec ^{2}$ (upward).

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