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Free fall acceleration problem

by COCoNuT
Tags: flowerpot
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COCoNuT
#1
Sep8-04, 04:57 PM
P: 35
A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes time 0.420 s to pass this window, which is of height 1.90 m.

Question- How far is the top of the window below the windowsill from which the flowerpot fell? Take the free fall acceleration to be 9.80m/s^2 .


given:
t = 0.420s
h = 1.90m

using the position formula

x(t) = x(0) + v(0)t + 1/2at^2
x(t) - x(0) = v(0)(0.42) - 4.9(0.42)^2
1.9 = v(0)(0.42) - 4.9(0.42)^2
v(0) = 6.59m/s

vf^2 - v(0)^2 = 2gh <--- now using this formula....
(6.582)^2 - 0 -2(-9.8)h
h = (6.582)^2/(9.8^2)

h = 2.21m

my answer is 2.21m, but it's wrong. i know im close, can someone help?
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bullet_ballet
#2
Sep8-04, 05:09 PM
P: 15
Let u be the distance between the windowsill and the top of the window. If you set the windowsill at y=0, then use the formula [tex]y=y_0-v_0t-\frac{1}{2}gt^2[/tex]. We have y0=0 by our reference point and v0=0 since the flowerpot falls. Therefore:

[tex]y=-\frac{1}{2}gt^2[/tex]
[tex]-1.9-u=-\frac{1}{2}gt^2[/tex]
[tex]u=\frac{1}{2}gt^2-1.9[/tex]
Doc Al
#3
Sep8-04, 06:34 PM
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Quote Quote by COCoNuT
using the position formula

x(t) = x(0) + v(0)t + 1/2at^2
x(t) - x(0) = v(0)(0.42) - 4.9(0.42)^2
1.9 = v(0)(0.42) - 4.9(0.42)^2
v(0) = 6.59m/s
Your thinking is correct, but you are not using a consistent sign convention. Since you are taking the acceleration to be -9.8 m/s^2, that means you chose down as negative, up as positive. But you weren't consistent. x(t) - x(0) = -1.9 (since the change in position is negative), which makes the equation:
-1.9 = v(0)(0.42) - 4.9(0.42)^2; solve this one for v(0). (And realize that v(0) will, of course, be negative!)

vf^2 - v(0)^2 = 2gh <--- now using this formula....
(6.582)^2 - 0 -2(-9.8)h
h = (6.582)^2/(9.8^2)
Of course, you need to plug in the new, correct value for v(0). But use that same sign convention. Note that a is negative, so h will turn out to be negative. Which makes sense in your sign convention, since the top of the window is below the starting point.

FYI: No one says you must choose down to be negative. You could call down positive, in which case a = +9.8 m/s^2, etc. Just be consistent and you'll do fine.

CellCoree
#4
Sep8-04, 09:23 PM
P: 42
Free fall acceleration problem

Quote Quote by Doc Al
Your thinking is correct, but you are not using a consistent sign convention. Since you are taking the acceleration to be -9.8 m/s^2, that means you chose down as negative, up as positive. But you weren't consistent. x(t) - x(0) = -1.9 (since the change in position is negative), which makes the equation:
-1.9 = v(0)(0.42) - 4.9(0.42)^2; solve this one for v(0). (And realize that v(0) will, of course, be negative!)


Of course, you need to plug in the new, correct value for v(0). But use that same sign convention. Note that a is negative, so h will turn out to be negative. Which makes sense in your sign convention, since the top of the window is below the starting point.

FYI: No one says you must choose down to be negative. You could call down positive, in which case a = +9.8 m/s^2, etc. Just be consistent and you'll do fine.
coconut, i have the same problem as you(with different numbers), are you doing your hw on mp also?


-1.9 = v(0)(0.42) - 4.9(0.42)^2 after solving this for v(0) i get -2.46m/s

(-2.46)^2 =2(-9.8)h
h = -.308m
does that look right? seems incorrect to me
bullet_ballet
#5
Sep8-04, 10:07 PM
P: 15
If the flowerpot is just falling from rest and not being thrown or propelled downwards, then v0 will equal zero.
CellCoree
#6
Sep8-04, 10:16 PM
P: 42
Quote Quote by bullet_ballet
If the flowerpot is just falling from rest and not being thrown or propelled downwards, then v0 will equal zero.
didnt i set v(0) to zero for the second equation?

can someone tell me if h = -.308m is correct?
bullet_ballet
#7
Sep8-04, 11:25 PM
P: 15
Using the equations in my first post, I get 1.03m as the distance between the sill and the window.
CellCoree
#8
Sep8-04, 11:31 PM
P: 42
Quote Quote by bullet_ballet
Using the equations in my first post, I get 1.03m as the distance between the sill and the window.
yea i did use that equation,

-1.9 = v(0)(0.42) - 4.9(0.42)^2

solved for v and plugged it into

vf^2 - v(0)^2 = 2gh

dont know why i get a different answer


i even used your formula and get -2.764
bullet_ballet
#9
Sep8-04, 11:38 PM
P: 15
Quote Quote by CellCoree
yea i did use that equation,

-1.9 = v(0)(0.42) - 4.9(0.42)^2

solved for v and plugged it into

vf^2 - v(0)^2 = 2gh

dont know why i get a different answer


i even used your formula and get -2.764
My bad. I had the wrong signs in there. Since I set y=0 at the windowsill, all values should be negative. Therefore, if -y is the distance to the bottom of the window from the windowsill, it should be broken up into -u-1.9 where u is defined as before. Therefore, [tex]-u-1.9=-\frac{1}{2}gt^2[/tex]. I made the chages to the above equations as well.
Doc Al
#10
Sep9-04, 07:58 AM
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P: 41,323
Quote Quote by CellCoree
-1.9 = v(0)(0.42) - 4.9(0.42)^2 after solving this for v(0) i get -2.46m/s

(-2.46)^2 =2(-9.8)h
h = -.308m
does that look right? seems incorrect to me
Looks good to me. Be sure to reword it so you answer the exact question that was in the original problem. The question asked "How far is the top of the window below the windowsill from which the flowerpot fell?"--so I would say the answer is 0.31 m.


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