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.

Answers (1)

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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