Finding height dropped from Force vs. Time graph

AI Thread Summary
The discussion revolves around calculating the height from which a pumpkin was dropped based on data from a force vs. time graph. The key equations used include the conservation of energy and the relationship between impulse and momentum. The area under the force vs. time graph represents impulse, which was estimated using a Riemann Sum, yielding a value of approximately 49.72 Nm. This impulse was then used to determine the final velocity of the pumpkin, leading to a calculated height of about 17.29 meters. The participants express uncertainty about the corresponding floor in the skyscraper but confirm the calculations are reasonable.
drierplease
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Homework Statement



A pumpkin was dropped from a balcony of a skyscraper. The pumpkin happened to land on a sidewalk force sensor and the below data was collected. The pumpkin weighs 2.7kg.

a) From what height was the pumpkin dropped?

b) From what floor of the skyscraper was the pumpkin probably dropped?

upload_2016-11-29_13-40-39.png


Homework Equations



Potential Before = Kinetic After: mgh = (1/2)mv^2

Solving for v gives: v = sqrt(2gh)

a = F/m

Vf = a*t+Vi
Sf = .5*at^2+Vi*t+ Si

The Attempt at a Solution



First I used mgh = .5mv^2 and solved for v.

This gave me v = sqrt(2gh)

I figure that finding the area under the curve of the graph gives the total force on the pumpkin as it is hitting the ground. I haven't done this yet, but I am not too worried about getting an exact answer for this part. I more want to make sure I know how to solve the rest of the problem.

I tried using kinematic equations to solve for height, but I ended up with h being 0, so I think I am not using the right concept and/or equation here.

We are currently studying conservation of momentum, but I don't see how that is helpful here for determining the height the object was initially dropped from.

Hopefully I am missing something obvious?
 
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Hello. Welcome to PF!

The area under the graph does not give you the "total force". It gives you something called "impulse". Review your notes or textbook for information about impulse and how it is related to momentum.
 
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TSny said:
Hello. Welcome to PF!

The area under the graph does not give you the "total force". It gives you something called "impulse". Review your notes or textbook for information about impulse and how it is related to momentum.

Right, I was mistaken. I think I've got it now.

The impulse is equal to the change in momentum over that time period.

Change in momentum = area under the curve = impulse = J

So, using that I solved change in momentum:

2.7(v) - 2.7(0)
2.7v = J
v = J/2.7

I estimated J by using Reimann Sum, got 49.72.

Then I just used conservation of energy, mgh = .5*m*v^2 and got 17.29 meters for h.

Hopefully this is right, I think I didn't quite understand impulse.
 
drierplease said:
I estimated J by using Reimann Sum, got 49.72.

Then I just used conservation of energy, mgh = .5*m*v^2 and got 17.29 meters for h.
I got J = 49.65 Nm and h = 17.24 m.
I'm not sure what floor that was from, but when thinking "skyscraper", it was not very high.
Once again, a problem with more drama and suspense needed. :)
 
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TomHart said:
I got J = 49.65 Nm and h = 17.24 m.
I'm not sure what floor that was from, but when thinking "skyscraper", it was not very high.
Once again, a problem with more drama and suspense needed. :)

Thanks, I'm glad I was close!
 
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