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Question

An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height of 100 m. What is the difference in their heights after 2 seconds if both the objects drop with the same accelerations? How does the difference in heights vary with time?


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Solution

Given:

The height of the first object, h1=150m

The height of the second object, h2=100m

Finding the height d1 of the first object after 2 seconds:

Let s1 be the distance covered by the first body in 2 seconds. Then,

d1=h1s1

by using the equation of motion, s=ut+12gt2

s1=0×2+12gt2Asu=0forafreelyfallingobjects1=12gt2Similarly,s2=12gt2Now,findingtheheightofthefirstobjectusing,d1=h1s1d1=150(12gt2)d1=150(12×10×4)d1=15020d1=130m

Therefore the height of the first object after 2 seconds is 130 m.

Finding the height d2 of the second object after 2 seconds:

Let s2 be the distance covered by the second body in 2 seconds. Then,

d2=h2s2

d2=100(12gt2)d2=100(12×10×4)d2=10020d2=80m

Therefore, the height of the second object after 2 seconds is 80 m.

Finding the difference between the heights of the two objects:

d1d2=130m80md1d2=50m

So, the difference is the same, i.e. 50 m.

This concludes that the difference in height of the two objects does not depend on time and will always be the same.


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