A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
Answers
Answered by
50
Explanation:
Given, acceleration of stone is a=g=9.8m/s
2
Initial velocity is u=0m/s
Time for hitting water is t=3.41s
If d be the depth of the well.
Using d=ut+at
2
/2,
∴ d=0(3.41)+0.5(9.8)(3.41)
2
=56.98∼57m
Answered by
57
Answer:
- Depth of well = 56.98 metres [approx.]
Explanation:
Given that,
Stone is dropped into a deep well, and is heard to hit the water after 3.41 s of being dropped.
so,
- Initial velocity of stone, u = 0
- acceleration due to gravity, a = 9.8 m/s²
- time taken by stone to reach the water surface, t = 3.41 s
We need to find,
Depth of well
that is,
- Distance covered by stone, s =?
Formula required,
- Second equation of motion
s = u t + 1/2 a t²
[ Where s = distance covered; u = initial velocity, a = acceleration; t = time taken ]
Calculation,
Using second equation of motion
→ s = u t + 1/2 a t²
→ s = ( 0 ) ( 3.41 ) + 1/2 ( 9.8 ) ( 3.41 )²
→ s = ( 4.9 ) ( 3.41 ) ( 3.41 )
→ s = 56.98 m [approx.]
Therefore,
- Depth of the well is 56.98 metres approximately.
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