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A stone is dropped down from a deep well from rest . the well is 50m deep . how long will it take to reach the bottom of the well?

2
by kalps

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by kalps

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u=0 m/s

S=50 m

g=9.8 m/(s^2) (for freely falling body,g is +ve)

t=?

For a freely falling body dropped from rest,

S=1/2g*t^2 (S=ut+1/2a*t^2=>S=0+1/2g*t^2=>S=1/2g*t^2)

50=1/2*9.8*t^2

t^2=50*2/9.8

t^2=10.204

=>t=(10.204)^1/2

=>t=3.19 sec

hence,it will take 3.19 sec to reach bottom of the well.

We know that ,

v=0.5*gt

s/t=0.5*gt

s=0.5*g*t*t

50=0.5*9.8*t*t

t^2=50/4.9

t=3.194

v=0.5*gt

s/t=0.5*gt

s=0.5*g*t*t

50=0.5*9.8*t*t

t^2=50/4.9

t=3.194