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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. Calculate when and where the two stones will meet.
Case(i): When the stone dropped from the top of the tower,
Let the distance travelled in time (t) = x
Using the second equation of motion,
Case(ii): When the stone is thrown upwards,
Initial velocity u = 25 m/s
Distance travelled in time (t) = (100-x)
From equations (1) and (2), we get
After 4sec, two stones will meet.
Now, by substituting the value of t in equation (1), we get
Putting the value of in , we get,
(100-80) = 20 m.
This means that after 4sec, 2 stones will meet at a distance of 20 m from the ground.
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