Question

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is
projected vertically upwards from the ground with the velocity of 25m/s. Calculate when and where the stons will meet . (take g= 10m/s^2​

Answer 1

SOLUTION

Let the position is y meter below the top.

So distance travelled by the stone thrown upward is (100 - y)m

Time taken by the stone to cover these distances is same.

• For first stone.

Time taken to cover this distance = t

Initial velocity = 0

acc to eq of motion.

y = ut + (1/2)gt²

y = 5t²

(y/5) = t²

y = 5t²

• For second stone

Time taken is same

Initial velocity = 25m/sec

100 - y = 25t - 5t².......put y = 5t² here

100 - 5t² = 25t - 5t²

25t = 100

time = 4sec

We know that y = 5t²

Y = 5(4)²

y = 5(16)

Y = 80m.....from the top

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