Given data:
(i) When the stone from the top of the tower is thrown,
Initial velocity u = 0
Distance travelled = x
Time taken = t
Therefore,
\( s = ut + \frac{1}{2}gt^2\)
x = 0 + (1/2)gt2
x = 5t2 ____________(a)
(ii) When the stone is thrown upwards,
Initial velocity u = 25 m/s
Distance travelled = (100 – x)
Time taken = t
\( s = ut + \frac{1}{2}gt^2\)
\((100-x) =25t + (1/2)\times10xt^2\)
\( x =100-25t + 5t^2 \)----------(b)
From equations (a) and (b)
5t2 = 100 -25t + 5t2
t = (100/25) = 4sec.
After 4sec, two stones will meet
From (a)
x = 5t2 = 5 x 4 x 4 = 80m.
Putting the value of x in (100-x)
= (100-80) = 20m.
This means that after 4sec, 2 stones meet a distance of 20 m from the ground
Answered by Shivani Kumari | 1 year ago
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