A 3.2 kg flowerpot drops from a tall building. The initial speed of the pot is zero, and you may neglect air resistance.(adsbygoogle = window.adsbygoogle || []).push({});

a) What is the magnitude of the force acting on the pot while it is in the air 1.5 s after it begins to fall?

b) After the pot has fallen 28 m, what is its speed?

c) After the pot has fallen 28 m, it enters a viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?

d) What is the force exerted on the liquid by the pot?

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Okay, I got the answers for the first three, which are

1) 31.36 N

2) 23.4 m/s

3) 182 m/s/s

If you really want to know how I did it, I can provide that in a later post, but thats not the point here.

I'm stuck on the last question.

I know I have to use this equation; F=M*a (in the y direction)

Well, I was thinking, the object has a downward force of 31.36 N, so when it hits the liquid, shouldn't the liquid push back up 31.36N? Of course, my answer doesn't match the computer, so insight appreciated.

*31.36N achieved by 3.2kg*9.8m/s/s

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# Falling flowerpotstuck on simple part

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