body is released from rest from a height. Find the
distance it falls through in (a) 1s, (b) 2s, and (c)
3s.
Answers
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0
Answer:
For A to B−
h=(a)(t
1
)+
2
1
gt
1
2
[2
nd
equation of motion.]
t
1
=
g
2h
..................(1)
From A to C−
2h=(0)(t
2
)+
2
1
gt
2
2
t
2
=
g
4h
..................(2)
Similarly, t
3
=
g
6h
........(3
)
Times taken from A to B=t
1
=
g
2h
Times taken from B to C=t
2
−t
1
=
g
4h
−
g
2h
=
g
2h
(
2
−1)
Time taken from C to D=t
3
−t
2
=
g
6h
−
g
4h
=
g
2h
(
3
−
2
)
∴ Ration =1:(
2
−1):(
3
−
2
)
answer is = t1:
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