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प्रश्न
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.
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उत्तर
Here, h = 100 m
Let the two stones meet after t seconds at point P, which is at a height x above the ground, as shown in the figure.
For stone 1,
u = 0, h = (100 - x) m,
a = g = 9.8 m/s2
From s = `ut + 1/2 at^2`
(100 - x) = `0 + 1/2 xx 9.8 t^2`
= 4.9t2 ...(i)
For stone 2,
u = 25 m/s, h = x,
a = - g = - 9.8 m/s2
From s = `ut + 1/2 at^2`
x = `25t + 1/2 (-9.8)t^2`
= 25t - 4.9t2 ...(ii)
Adding equations (i) and (ii)
100 - x + x = 25t
⇒ t = `100/25`
= 4 s
From equation (i),
100 - x = 4.9 × (4)2
100 - x = 78.4
x = 100 - 78.4
x = 21.6 m
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