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Question

A stone is dropped from the top of building and at the same time a second stone is thrown vertically upward from the bottom of the building with a speed of 20 ms−1. They pass each other 3 seconds later. Find the height of the building.
[Take g=10 m/s2]

A
40 m
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B
60 m
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C
65 m
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D
80 m
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Solution

The correct option is B 60 m
As per the question, both the stones will meet after 3 sec.
Let, total height of the building be H and the distance covered by dropped and thrown stone be H1 and H2 respectively, we have
H=H1+H2
So, from equation of motion we have
s=ut+12at2
H1=0(3)+12×10×(3)2
H1=45 m
also, H2=(20)(3)12×10×(3)2
H2=15 m
Hence, total height of the building is
H=45+15=60 m

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