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atroni [7]
3 years ago
6

A rock dropped down a well takes 1.8 s to hit the water. How far below the top of the well is the surface of the water

Physics
1 answer:
Brut [27]3 years ago
3 0

Answer: 15.87 m

From the equation of motion:

s=ut+\frac{1}{2}at^2

where, s is the distance traveled, u is the initial velocity, a is the acceleration and t is the time.

The rock free falls under gravity. Initial velocity, u=0 m/s, a=g=9.8m/s^2

It took t=1.8 s for rock to hit the water.

Substitute the values in the given equation:

\Rightarrow s=0+\frac{1}{2}9.8m/s^2\times(1.8s)^2=15.87 m

Hence, the water is 15.87 m below the top level of the well.

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Answer:

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For this exercise let's use Newton's second law

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In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

               

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3 years ago
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