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Free fall acceleration problemby COCoNuT
Tags: flowerpot 
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#1
Sep804, 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? 


#2
Sep804, 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_0v_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]1.9u=\frac{1}{2}gt^2[/tex] [tex]u=\frac{1}{2}gt^21.9[/tex] 


#3
Sep804, 06:34 PM

Mentor
P: 41,568

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!) 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. 


#4
Sep804, 09:23 PM

P: 42

Free fall acceleration problem
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 


#5
Sep804, 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.



#6
Sep804, 10:16 PM

P: 42

can someone tell me if h = .308m is correct? 


#7
Sep804, 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.



#8
Sep804, 11:31 PM

P: 42

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 


#9
Sep804, 11:38 PM

P: 15




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