A stone is dropped from the top of a tower of height 125 m. The distance tracelled by it during last second of its fall is (g=10ms-2)
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Answered by
11
hn =1/2×g(2n-1)
= 10/2 (2×1-1)
=5m
= 10/2 (2×1-1)
=5m
Answered by
56
it's a free fall motion, so initial velocity u = 0
so from the 2nd equation of motion,
s = 1/2at^2 (as u = 0)
taking a = g and s = 125m ,
total time of journey t = 5s.
now distance travelled in the last second i.e in the 5th second is,
s = u + a/2(2n - 1)
putting u = 0, a = g and n =5,
s = 45m
so the answer is 45 m.
so from the 2nd equation of motion,
s = 1/2at^2 (as u = 0)
taking a = g and s = 125m ,
total time of journey t = 5s.
now distance travelled in the last second i.e in the 5th second is,
s = u + a/2(2n - 1)
putting u = 0, a = g and n =5,
s = 45m
so the answer is 45 m.
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