a body falling from aminrate travels 40 m in last 2 sec.what is ht of minrate
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This problem can easily be solved with the help of equation of motion.
s = ut + 1/2 at^2
40 = 2u + 1/2.10.(2)^2
40 = 2u + 20
2u = 20
u = 10 m /s
Now,
v^2-u^2 = 2as
2.10.s=100
s = 5m
Hence the total distance traveled is 45M.
s = ut + 1/2 at^2
40 = 2u + 1/2.10.(2)^2
40 = 2u + 20
2u = 20
u = 10 m /s
Now,
v^2-u^2 = 2as
2.10.s=100
s = 5m
Hence the total distance traveled is 45M.
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