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A stone is dropped down a deep from rest. The well is 50m.deep. how long will it take to reach the bottom of well? (acc.=9.8m/s?
Verified Answer
A stone is dropped down a deep from rest. The well is 50m.deep. how lo...
Given  that,

a  =  9.8 m/s2

u  =  0

s  =  50m



By  Newton's  law  of  motion,

v2  -  u2  =  2as,

ie,  v2  -  [0]2  =  2 * 9.8 * 50

ie,  v2  =  980.

So,  v  =  Root  of  980

=  31.304....



Now,  we  got  the  final  velocity  v.

So,  we  can  find  the  time  required,  by  Newton's  equation  of motion,

v  =  u  +  at.

ie,  31.304  =  0  +  9.8 * t

31.304  =  9.8t

So,  t  =  31.304  /  9.8

ie,  31.30 * 100  /  9.8 * 100

=  3130  /  980

=  3.193....m / s.
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Most Upvoted Answer
A stone is dropped down a deep from rest. The well is 50m.deep. how lo...
Given, u = 0 m/s. u = Initial velocity v = ? v = Final velocity D = 50m. D = Distance A = 9.8m/s. A = Acceleration T = ? T = Time We know, v²-u² = 2as So, v²-0² = 2 × 9.8 × 50 v² = 980 v = 31.304...m/s TIME. S = ut + 1/2 at² 50 = 0t + 1/2 × 9.8 × t² 50 = 0 + 4.9t² 50 = 4.9t² 50/4.9 = t² 10.2 = t² 3.1937 sec = tHOPE IT HELPED 🙂
Community Answer
A stone is dropped down a deep from rest. The well is 50m.deep. how lo...
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.
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