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Physics-

General

Easy

Question

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

$2100 m / \sec^{2} d o w n w a r d s$
$2100 m / \sec^{2} u p w a r d s$
$1400 m / {s e c}^{2}$
$700 m / {s e c}^{2}$

The correct answer is: $2100 blank m divided by sec to the power of 2 end exponent invisible function application u p w a r d s blank$

Velocity at the time of striking the floor,
$u equals square root of 2 g h subscript 1 end subscript end root equals square root of 2 cross times 9.8 cross times 10 end root equals 14 m divided by s$
Velocity with which it rebounds
$v equals square root of 2 g h subscript 2 end subscript end root equals square root of 2 cross times 9.8 cross times 2.5 end root equals 7 blank m divided by s$
$therefore C h a n g e blank i n blank v e l o c i t y blank increment v equals 7 minus open parentheses negative 14 close parentheses equals 21 m divided by s$
$therefore A c c e l e r a t i o n equals fraction numerator increment v over denominator increment t end fraction equals fraction numerator 21 over denominator 0.01 end fraction equals 2100 m divided by s to the power of 2 end exponent left parenthesis u p w a r d s right parenthesis$

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