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