A body is dropped from rest at a height of 150m and simultaneously another body is dropped from rest from point 100 m above the ground. What is their difference in height after they have fallen (a) 2s and (b) 3s? How does the difference in height vary with time?
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Answer:
Initial difference in height =(150−100)m=50m
We know that, by second equation of kinematics, s=ut+
2
1
at
2
considering g=10m/s
Distance travelled by first body in 2s=h1=0+(
2
1
)g(2)
2
=2g=2×10=20m
Distance travelled by another body in 2s=h2=0+(
2
1
)g(2)
2
=2g=2×10=20m
After 2s, height at which the first body will be =h
1
=150−20=130m
After 2s, height at which the second body will be =h
2
=100−20=80m
Thus, after 2 s, difference in height =(130−80)
=50m=initial difference in height
Thus, difference in height does not vary with time. So the answer is zero.
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