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Question

A ball is dropped on the floor from a height of $$10$$ m. It rebounds to a height of $$2.5$$ m. If the ball is in contact with the floor for $$0.01$$ sec, then average acceleration during contact is


A
2100m/s2
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B
1400m/s2
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C
700m/s2
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D
400m/s2
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Solution

The correct option is C $$2100 m/s^2$$
Velocity of the ball on reaching the ground is 
$$v^2 = u^2 + 2as $$ (u is $$0$$)
$$v^2 = 2g(10) $$
$$v = 14 m/s $$
After striking the ground the ball bounces off. Velocity of the ball immediately after striking the ground is
$$v^2 = u^2 - 2gh $$ (v is zero at maximum height)
$$u^2 = 2g(2.5) $$
$$u = 7m/s $$
Time of impact on the ball $$= 0.01 s $$
avg. acceleration$$ = \dfrac{v_f - v_i}{t} $$
$$a =\dfrac{7 - (-14)}{0.01} $$ (since both velocities are in opposite directions.)
$$ a = 2100 m/s^2$$

Physics

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