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?
shashidharybhat:
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Answers
Answered by
7
We can use the equation s=ut+(1/2)at²
Where s is the distance(50m)
u is the initial velocity(0)
acceleration due to gravity=9.8m/s²
time taken
s=ut+(1/2)at²
50=(1/2)*9.8*t²
50=4.9t²
t²=50/4.9
t=√50/√4.9
t=3.19=3sec(approx)
Where s is the distance(50m)
u is the initial velocity(0)
acceleration due to gravity=9.8m/s²
time taken
s=ut+(1/2)at²
50=(1/2)*9.8*t²
50=4.9t²
t²=50/4.9
t=√50/√4.9
t=3.19=3sec(approx)
Answered by
3
s=50m
a=g=9.8m/sec^2
u=0
now, we know
s=ut+1/2at^2
50=(0)×t+1/2(9.8)t^2
50×2=(9.8)t^2
100/9.8=t^2
100×10/98=t^2
100×5/49=t^2
√100×√5/√49=t
t=10√5/7 sec
a=g=9.8m/sec^2
u=0
now, we know
s=ut+1/2at^2
50=(0)×t+1/2(9.8)t^2
50×2=(9.8)t^2
100/9.8=t^2
100×10/98=t^2
100×5/49=t^2
√100×√5/√49=t
t=10√5/7 sec
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