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Old 09-26-05, 01:45 AM   #1
ovoleg
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Help with physics problem? THanks! :)

IF anyone can help with this physics problem, I will owe you big time

I tried this problem with someone else and they couldn't figure it out either.

The answer is :.310m

"A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes .420s to pass this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell?"

I tried, x=x0+v0(t)+1/2(a)(t)^2. v=v0+at

...

Its some simple problem which is a PITA. I have the answer I just can't find a way to get to it.

THanks anyone
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Old 09-26-05, 03:16 AM   #2
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x=x+vt+0.5at^2 should be used - but remember that the starting velocity of the journey for which the numbers are given is not 0. You need to calculate what that velocity was, and then use that to get your answer.

journey1 = sill to top of window
journey2 = top to bottom of window


Journey 2
x2=bottom of window
x1=top of window
v=velocity at top of window
t=time to travel across window

x2=x1 + vt + 0.5at^2 => you get v=2.4658

(start velocity of journey 2 = final velocity of journey 1)...

Journey 1:

So now you can use this value of v to calculate the amount of time it took to get there: v=at => you get t=0.2516

This is the amount of time in journey1. Now you can subtitute directly into the old favourite...

x=0+0+0.5at^2 = 0.310
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Old 09-26-05, 03:24 AM   #3
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Here's how I did it, you have to set up the problem into multiple parts first. Draw a picture and it makes sense. You're trying to solve it all in one equation and that's not possible because you have to solve for multiple variables, thus multiple equations are necessary. Here's a picture:

The various equations are (assuming constant acceleration):

1.
average Velocity across window = dX/dT = 1.90m/0.420s = 4.524m/s


2.
average Velocity across window = (V1+V2)/2

Solving for V1 we get (V1+V2)/2 = 4.524m/s
V1 + V2 = 9.048m/s
V1 = 9.048m/s - V2


3.Then solving for acceleration across 1.90m we get:

average Acceleration across window = dV/dT = (V2-V1)/0.420s
V2 = V1 + at = V1 + (9.8m/s^2 * 0.420s) = V1 + 4.116m/s
V2 = V1 + 4.116m/s

Combining these last two equations we get:

V1 = 9.048m/s - V2
V2 = V1 + 4.116m/s
---------------------
V1 = 9.048m/s - (V1+4.116m/s)
V1 = 9.048m/s - V1 - 4.116m/s
2V1 = 9.048m/s - 4.116m/s = 4.932m/s
V1 = 2.466m/s is velocity at top of lower window.


4. Pluging into V=at for velocity over constant acceleration we get:
2.466m/s = 9.8m/s^2 * t
t = 0.252s time to fall to top of lower window


5. Pluging into d=1/2at^2 for distance over time w/constant acceleration we get:
d = 0.5at^2 = 0.5 * 9.8m/s^2 * 0.252s^2

d = 0.310m (don't round off numbers until final step)

Last edited by DannoXYZ; 09-26-05 at 06:33 PM.
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Old 09-26-05, 08:16 AM   #4
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....depends on what species of plant it is
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Old 09-26-05, 09:34 AM   #5
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omg thanks so much guys!!! I think I put in 2hrs trying to solve this. I wasted the first hour because I misread the question, I thought the total distance traveled took .42s.

Thanks guys!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Old 09-26-05, 02:25 PM   #6
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Why did I even open this thread...it is soooo over my head. I think a flowerpot landed on my head in college...or maybe it was what I was smokin that was IN the flowerpot!
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Old 09-26-05, 02:30 PM   #7
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After I got the right problem down, I was thinking with the same sense as Danno but I didn't set up the right equations. Thanks so much, I actually got to a point where I thought the answer was 1.03 and thought the teacher typo'ed and meant 1.03 instead of .310 because .31 is only the significant figure and to write .310 doesn't make any sense.

Thanks again!
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Old 09-26-05, 11:09 PM   #8
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This can all be avoided if you stick with ground floor appartments.
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