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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.

Q. 17.     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.

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Let the distance travelled by the stone which is dropped from the top upto the instant when the two stones meet be x

Initial velocity u = 0

Acceleration a = g = 9.8 m s-2

Using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ x=0\times t+\frac{1}{2}\times 9.8\times t^{2}\\ x=4.9t^{2}$$(i)$

The distance travelled by the stone which is projected vertically upwards from the ground up to the instant when the two stones meet would be equal to 100 - x

Initial velocity = 25 m s-1

Acceleration a = -g = -9.8 m s-2

Using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ 100-x=25\times t+\frac{1}{2}\times (-9.8)\times t^{2}\\ 100-x=25t-4.9t^{2}\\ x=4.9t^{2}-25t+100$$(ii)$

Equating x from (i) and (ii) we get

4.9t2 = 4.9t2  - 25t + 100

25t = 100

t = 4s

x = 4.9t2

x = 4.9 X 42

x = 78.4 m

100 - x = 100 - 78.4 = 21.6 m

The stones meet after a time of 4 seconds at a height of 21.6 metres from the ground.

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