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.

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

We know that, $$s = ut + \frac{1}{2}gt^2 \ => \ x = 0 + \frac{1}{2}gt^2 \ => \ x = 5t^2$$        - I

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

We know that, $$s = ut + \frac{1}{2}gt^2 \ => \ 100-x = 25t + \frac{1}{2} × 10 × t^2 \ => \ x = 100 - 25t + 5t^2$$        - II

From equations I and II

$$5t^2 = 100 -25t + 5t^2$$

=> $$t = \frac{100}{25} = 4 sec$$

So, after 4 sec, two stones will meet

Now, from I

$$x = 5 t^2 = 5 × 4 × 4 = 80 m$$

Putting the value of x in (100-x), we get

$$x = 100-80 = 20 m$$