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Freefall- kinematics problem.

  • #1
QuantumCurt
Education Advisor
726
166

Homework Statement



A cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of .50s and the top-to-bottom height of the window is 2.00m. How high above the window does the flower pot go?


I was trying to help one of my friends with this problem, and apparently I'm not setting it up right. I can't figure out what I'm doing wrong.



The Attempt at a Solution



I figured the velocity would be 8.0 m/s, because it's in view for a total of .50 seconds, between the two passes that it is making by the window.

So, one pass is 2.00m/.25 s=8 m/s

Then I used a kinematic to solve for displacement.

[tex]v^2=v_{0}^2+2a\Delta{x}[/tex]


Now I let the final velocity be 0 at the top of the flight path, and I substitute g in for a, with downward being the negative direction. Then solve for delta x, and substitute numerically.

[tex]\Delta{x}=\frac{-v^2_{0}}{-2g}[/tex]

[tex]\Delta{x}=\frac{-64}{-2(9.80)}[/tex]

Which gives me [itex]\Delta{x}=3.3 m[/itex] abpve the window.

Her book apparently says the answer is 2.34 m. Where did I go wrong?

I feel like the velocity calculation is kinda questionable here. I can't think of how else to go about it though. I tried a couple different approaches, and came up with answers that were much farther off.
 

Answers and Replies

  • #2
6,054
390
So, one pass is 2.00m/.25 s=8 m/s
If the flowerpot is in free fall, then this is wrong.
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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You are assuming that the velocity of the pot is constant as it passes by the window. Since the pot is affected by gravity going up as well as coming down, this assumption is invalid.
 
  • #4
QuantumCurt
Education Advisor
726
166
How would I go about calculating it? It's going upward initially, and I factored in the fact that gravity is slowing it down. I've been wracking my brain over this one for a while now, and I honestly cannot figure it out.
 
  • #5
QuantumCurt
Education Advisor
726
166
Can I use

[tex] v=v_{0}+at[/tex]

to find the time, then use

[tex]\Delta{x}=v_{0}t+\frac{1}{2}at^2[/tex] to find the height?

That's the only thing I can think of here. I think that was one of the things I tried though, and it wasn't working.
 
  • #6
6,054
390
Can I use

[tex] v=v_{0}+at[/tex]

to find the time
You do not have to find the time, it is already given to you. Note, however, than in this equation both ##v## and ##v_0## are unknown, so you cannot find anything from this equation alone.

[tex]\Delta{x}=v_{0}t+\frac{1}{2}at^2[/tex] to find the height?
The height is known. So are the time and the acceleration. The only unknown here is ##v_0##, which means you can find it. Once you get it, can you solve the rest of the problem?
 
  • #7
QuantumCurt
Education Advisor
726
166
You do not have to find the time, it is already given to you. Note, however, than in this equation both ##v## and ##v_0## are unknown, so you cannot find anything from this equation alone.



The height is known. So are the time and the acceleration. The only unknown here is ##v_0##, which means you can find it. Once you get it, can you solve the rest of the problem?
I got it. This was a really simple problem, I don't know why it wasn't clicking. Thanks for the help!!
 

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