Question

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?

Answer 1

Given,
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.

Answer 2

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