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 i to be uniform.
Answers
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Answer:
Consider the motion of ball from A to B.
B→ just above the sand ( just to penetrate)
u=0,a=9.8 m/s
2
,s=5 m
The distance covered by the ball is:
S=ut+
2
1
at
2
⇒5=0+
2
1
(9.8)t
2
⇒t
2
=
4.9
5
=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
u
1
=9.89 m/s,v
1
=0,a=?,s=10 cm=0.1 m
a=
2s
v
1
2
−u
1
2
=
2×0.1
0−(9.89)
2
=−490 m/s
2
The retardation in sand is 490 m/s
2
.
Explanation: