# Falling pot

1. Feb 15, 2015

### sorressean

Hello all:
I have a question. I'm looking at this problem:
1. The problem statement, all variables and given/known data
A flower pot falls past your window. The pot is visible for a time t, and the vertical length of your window is L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

From what height h above the bottom of your window was the flower pot dropped?

2. Relevant equations
I know these variables:
a = 9.8
x_i = 0

3. The attempt at a solution
I know that I can use the bottom of the window as x_f. I'm not really sure how to start out here though, any information would be really useful.
Thanks,

2. Feb 15, 2015

### Borg

Are you familar with calculating the velocity at a given time?

3. Feb 15, 2015

### sorressean

4. Feb 15, 2015

### sorressean

Hello:
I am. you use v=v0+at, correct?
So, a = 9.81.
t is unknown with the exception of the time it takes the pot to fall past the window.
v0 is just 0 starting out as it's just dropped.
Sorry for the double post. I'm blind and learning how to use this with a screen reader. didn't mean to quote myself.

5. Feb 15, 2015

### Borg

OK, I'll try not to quote too much.
Sorry, I meant to say the distance equation: d = vt + 1/2at2

6. Feb 15, 2015

### sorressean

Hello:
It's my fault. quoting is fine, I just replied to the quote and didn't edit for some reason.

yes, I am familiar with that equation.
So, in this case:
D is the displacement: the point the pot is dropped to the bottom of the window.
v is 0 to begin with.
The only t we know is the time it takes for the pot to fall l_w. so the distance is x+l_w (height of the window)
a is 9.8
t is unknown.
I'm still really confused (sorry)

7. Feb 15, 2015

### Borg

t is known - it is the time that it takes the pot to pass the window. The question wants you to describe the answer in terms of t and the window height Lw.

8. Feb 15, 2015

### Borg

I've got to go but someone else may jump in. For now, think about how you can apply the velocity and distance equations at time t1 and t2 which are the times when the pot crosses the top and bottom of the window. Keep in mind that T = t2 - t1 where T is the time that it takes to pass the window.

9. Feb 15, 2015

### haruspex

You need to break the problem into two stages: from initial release to top of window, and from there to bottom of window.
The information given all relates to the second stage, so deal with that first. You know the distance, the time, and the acceleration. There are two things that allows you to calculate. Which one will be useful for the first stage?

10. Feb 15, 2015

### sorressean

Hello:
Okay, here's my attempt then. Please let me know where I'm going wrong. I'm not sure how to handle some of this.
d+l_w = vt + 1/2a(w_t+t)^2
Thanks a lot for the help.

11. Feb 15, 2015

### haruspex

Is that in response to my post? That's not dealing with the second stage in isolation. What are the distance, time and acceleration for the second stage? What two things can you calculate directly from there? (I.e., which two SUVAT equations contain distance, time and acceleration?)

12. Feb 15, 2015

### azizlwl

As haruspex noted, you should break into segments. Pot drops to top of window. Second, length of window.

13. Feb 17, 2015

### sorressean

Hello:
I'm sorry, I missed your post initially.
So for the window, you already know l_w (height of window), t (time it takes to pass) and a:
l_w = v0 + 1/2*9.81*t^2
where t is the time it took to go l_w distance.
so you also need a velocity, which you can get:
v_f = v_i + at.
so this is from drop to the top of the window.
v_f = 0+9.81t
can you now factor this into the equation above?
l_w = (0+9.81t)+ 1/2*9.81*t^2
Thanks again for the help,

14. Feb 17, 2015

### haruspex

You've missed something out of that equation. Just a typo?
No, you can solve that equation to find v0. L_w and t are given. (Don't write 9.81, leave it as g. This is an algebraic exercise, not a numerical one.)
Careful - this is a different t. This vf is the v0 of the previous equation. You don't care how long it took the pot to reach top of the window, so look for an equation that does not involve time for this stage.

15. Feb 17, 2015

### sorressean

I feel like I'm just running in circles taking random jabs at this; are you able to offer anything more concrete? The equations I gave are the only things I have. I don't have a mysterious equation with no time that might work.

16. Feb 17, 2015

### Elvis 123456789

Since you do not have enough information to solve for "h" right away, you must use the other information given to you in order to deduce enough information about "h". You are given the time (t), the acceleration (g), and the distance L_w, for when the pot was falling past the window. Use this information to solve for the initial velocity within this interval ( the top of the window to the bottom) , which will in turn be the final velocity of the first interval, which is "h". Then use the kinematic equation that relates initial and final velocities, and solve for the displacement, which is "h".

17. Feb 17, 2015

### haruspex

Please post the corrected version of your l_w = v0 + 1/2*9.81*t^2, using g instead of 9.81, and rearranging it into the form v =...
There are five standard SUVAT variables. ("SUVAT refers to the equations that apply under constant acceleration.) They are two speeds (initial and final), distance covered, acceleration, and time elapsed. There are five equations. Each equation involves four of the five variables, so if you know any three you can calculate the other two.
You ought to know all five equations and be able to select the one that does not involve time. (Hint: it's the same as the work conservation equation but with mass divided out.)

18. Feb 17, 2015

### lightgrav

you wrote L = v0 + ... without the "t" .
there are 3 interesting time instants:
t0 = time it is dropped (with v0 = 0)
ttop = time at window top ; tbottom = time at window bottom
there is one early time duration, from the drop to window top,
and one later time duration from window top to window bottom.