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Question

# A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. The two stones will meet at a height

A
21.6 m from the ground
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B
50.0 m from the ground
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C
78.4 m from the ground
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D
19.6 m from the ground
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Solution

## The correct option is A 21.6 m from the groundLet u1 and u2 be the initial velocities of the stone projected from top and bottom respectively. Taking upward as positive, u1=0 and u2=25 m/s u21=25 m/s S21=100 m arel=g−g=0 m/s2 Using second equation of motion, we get S21=u21t+12arelt2 100=25t ⇒t=4 s Time taken by the two stones to meet =4 s Height from the ground at which two stones meet is given by h=u2t−12×g×t2 h=25×4−12×9.8×42 ⇒h=21.6 m

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