An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?
Answers
Answer:
Step-by-s In the figure given below, find the value of x
tep explanation:
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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