A ball is dropped on to the floor from a height of 10 cm. It rebounds to a height of 2.5m. If the ball is in contact with the floor for o.o1 seconds , what is the average accelerationduring contact ??

Asked by SIDDHARTH SHARMA | 24th Nov, 2014, 11:12: AM

Expert Answer:

The height from which the ball is dropped cannot be 10 cm, it will be 10 m as the ball is said to rebound 2.5 m.
Given, the height from which the ball is dropped, h1 =10 m,
The height to which the ball rebounds, h2 = 2.5 m 
Time of contact of the ball with floor, Δt = 0.01 sec
Let v1 be the velocity of the ball when dropped before striking the floor,
v2 be the velocity of ball upwards after striking the floor during rebounding
Change in velocity of ball during contact =v2 - (-v1)
begin mathsize 14px style Acceleration space straight a space equals fraction numerator straight v subscript 2 minus left parenthesis minus straight v subscript 1 right parenthesis over denominator triangle straight t end fraction space minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 1 right parenthesis When space the space body space is space dropped space from space height space straight h subscript 1 : straight u equals 0 comma space straight v space equals straight v subscript 1 space comma space straight a space equals straight g comma space straight S equals straight h subscript 1 We space know space the space equation space of space motion : space straight v to the power of 2 space end exponent equals straight u squared plus 2 aS straight v subscript 1 squared equals 0 plus 2 gh subscript 1 rightwards double arrow straight v subscript 1 equals square root of 2 gh subscript 1 end root space minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 2 right parenthesis Consider space the space motion space of space ball space after space striking space the space floor : straight u equals straight v subscript 2 space comma straight v space equals 0 space comma straight a space equals minus straight g comma space straight S space equals straight h subscript 2 Here space the space equation space of space motion space becomes : space 0 space equals straight v subscript 2 squared plus 2 cross times minus straight g cross times straight h subscript 2 rightwards double arrow straight v subscript 2 equals square root of 2 gh subscript 2 end root space minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 3 right parenthesis Substituting space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis space in left parenthesis 1 right parenthesis space we space get : straight a space equals fraction numerator square root of 2 gh subscript 2 end root plus square root of 2 gh subscript 1 end root over denominator triangle straight t end fraction Substituting space the space values space we space get : straight a space equals fraction numerator square root of 2 cross times 9.8 cross times 2.5 end root plus square root of 2 cross times 9.8 cross times 10 end root over denominator 0.01 end fraction equals 2100 space straight m divided by straight s squared end style
Average acceleration = 2100 m/s2

Answered by Jyothi Nair | 24th Nov, 2014, 03:36: PM

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