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.


Answer :

(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,
\( s = ut + \frac{1}{2}gt^2 \ \)
\( \Rightarrow \ x = 0 + \frac{1}{2}gt^2 \ \)
\( \Rightarrow \ x = 5t^2 \) ......- I

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

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


From equations I and II

\( 5t^2 = 100 -25t + 5t^2 \)

\( \Rightarrow t = \frac{100}{25} = 4 sec \)

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

\( x = 100-80 = 20 m \)

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