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A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically when and where the two stones will meet.
Case-I
In order to fall, h1 = ut + 0.5 at2
h1 = 0 x t + 0.5 gt2 ...(i)
Case-II,
In order to throw up,
h1 = ut - 0.5 at2
h1 = 25 x t + 0.5 gt2 …(ii)
from eq(i) and (ii)
h1 + h2 = 25t
100 = 25t
t = 100/25
t = 4s
Therefore the two stones will meet after 4s.
∴ Putting t = 4s in equation (i)
h1 = 0 x 4 + 0.5x 9.8 x 4 x 4
h1 = 0 + 78.4
h1 = 78.4 m
Therefore both the stone will meet at 78.4m from the top or (100 – 78.4) = 21.6 m from the bottom.
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