A stone is allowed to fall from the top of a tower 100 m
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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Let the stones meet at point A after time t.
For upper stone :
u′ = 0
x = 0 + 1/2 gt^2
x = 1/2 x 10 x t^2
⟹x = 5t^2 ............(1)
For lower stone :
u = 25 m/s
100 − x = ut − 1/2gt^2
100 − x = (25)t − 2 1 × 10 × t^2
⟹100 − x = 25t − 5t^2 ............(2)
Adding (1) and (2), we get
25t = 100
⟹ t = 4 s
From (1),
x = 5 × 4^2
⟹x=80 m
Hence the stone meet at a height of 20 m above the ground after 4 seconds.
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