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Question

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

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Solution

Consider the motion of ball from A to B.
B just above the sand ( just to penetrate)
u=0,a=9.8 m/s2,s=5 m

The distance covered by the ball is:
S=ut+12at2
5=0+12(9.8)t2

t2=54.9=1.02
t=1.01.

velocity at B,v=u+at=9.8×1.01(u=0)=9.89 m/s.

From motion of ball in sand
u1=9.89 m/s,v1=0,a=?,s=10 cm=0.1 m

a=v21u212s=0(9.89)22×0.1=490 m/s2

The retardation in sand is 490 m/s2.

1291740_1124188_ans_7ecc13735fdc4edead88dcd095110574.png

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