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Question

A stone is dropped from a tower of height 100 m. At the same time, another stone is thrown upwards at a speed of 25 m s1. When and where the two stones would meet? Take g=10 m s2

A
Time =2.5 s ; Height =40 m
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B
Time =4 s ; Height =80 m
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C
Time =6 s ; Height =50 m
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D
Time =7.5 s ; Height =100 m
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Solution

The correct option is B Time =4 s ; Height =80 m
Given:
The height of the tower, h=100 m
Initial velocity of the second stone, u2=25 m s1
g=10 m s2
Initial velocity of the first stone, u1=0 m s1
Let t be the time instant at which the two stones would meet/hit each other.
For the falling stone, the distance covered:
s1=u1t+12gt2
s1=0+12×10t2=5t2 ------(1)
For the rising stone, the distance covered:
s2=u2t12gt2
s2=25t12×10t2=25t5t2
But, s1+s2=100
5t2+25t5t2=100
25t=100t=4 s ----- (2)
Substitute (2) in (1):
s1=5×(4)2=5×16=80 m --- (3)
The two stones would meet each other at t=4 s, and at a height of 80 m.

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