# Homework Help: 1D Kinematics question

1. Nov 14, 2015

### David112234

1. The problem statement, all variables and given/known data
A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 m high.

Part A
How far is the top of the window below the windowsill from which the flowerpot fell?

2. Relevant equations
all kinematic equations
V=V_i + at^2
x=x_i + v_i(t) + .5 a(t)^2
V^2=V_i^2 + 2a (x-x_i)
x-x_i = (V_i + Vx)(t)(.5)

note, _i means initial

a= -9.802

3. The attempt at a solution
my free body diagram

|
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|_ x=1.90 + u V=0
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|_ x=1.90 V=? T=0
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|
|
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|_ x= 0 V=? T= .420

I tried the second formula for final X position to find V at T=0

0= 1.90+V? (0) + (.5)(-9.802)(0)^2
and I get
0=1.90 which is not correct
I cant use any any other formulas because I dont have the variables
what am I supposed to do, 1.90 is not the correct answer as I tried.
aslo I know someone else posted the same question from mastering physics, but I did not understand it well and I wouldl ike ot start from scratch

2. Nov 14, 2015

### CWatters

I don't follow/agree with your drawing. The pot starts on a windowsill and falls an unknown distance until it reaches the top of another window. It then falls another 1.9m in 0.42 seconds.

The pot passes the top of the window at some unknown "initial velocity". What equation can be used to find this initial velocity given the displacement (1.9m), the time (0.42) and the acceleration?

Then there is another equation that can be used to work out the answer to the problem.

3. Nov 14, 2015

### David112234

Why do you disagree with my diagram? I chose the bottom of the lower window as 0, going up from that is 1.90, the height of the window, and higher is the height of the window + what ever distance is between them, U.
For V, it is 0 when the pot is standing on the upper window, when it falls its Velocity is unknown
And for T is it took .420 seconds to fall the height of the window then T=0 is when it starts falling that distance

and for the formula, the second kinematics equation, I tried it and it didn't work out,

I tried the second formula for final X position to find V at T=0

0= 1.90+V? (0) + (.5)(-9.802)(0)^2
and I get
0=1.90 which is not correct

did I plug in the wrong values?

4. Nov 14, 2015

### PeroK

Given that the flower pot started at the first window and is falling downwards, it's illogical to choose 0 as the bottom of the second window. You ought to choose 0 where the flowerpot started. Otherwise, you're liable to get your equations and variables mixed up!

5. Nov 14, 2015

### David112234

shouldnt the problem still work out? only the negatives will change

6. Nov 14, 2015

### PeroK

I would make life as easy as possible. It hasn't worked out. I always follow these principles:

1) Understand the problem.

2) Set things up as simply as possible.

3) Solve.

I would definitely set $x = 0$ where the object starts and have all motion and accleration positive in the downward direction.

7. Nov 14, 2015

### David112234

|
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|_ x=0 V=0
|
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|_ x=? V=? T=0
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|
|
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|_ x= 1.90 V=? T= .420

like this?

8. Nov 14, 2015

### PeroK

I can see two ways to do this. There's the conventional way. In that case, you'd definitely have $T=0$ at start.

There's another way to do it, where $T = 0$ where you have it would work.

Also, question marks are not much use. Instead use symbols for the unknowns, so you can use them in equations.

I would have:

$x = 0, T = 0, V = 0$

$x = d, T = t_1, V = v_1$

$x = d + 1.9, T = t_1 + 0.42, V = v_2$

9. Nov 14, 2015

### David112234

alright, Il use that, so now use a formula to solve for d? but there is too many variables to be able to solve for it

10. Nov 14, 2015

### CWatters

Your original diagram has x=190 in two places. That's why I thought it wrong. Please read my post #2 again. There are enough known variables to solve it.

11. Nov 14, 2015

### CWatters

Oh sorry I see you have x=190+u at the top. My mistake you diagram was OK. But still see the rest of #2

12. Nov 14, 2015

### CWatters

Perhaps remind yourself of the SUVAT equations?

13. Nov 14, 2015

### David112234

I tried your suggestion, it does not help since the T is 0, or should I use t= .420?

Last edited: Nov 14, 2015
14. Nov 14, 2015

### David112234

Ok so I tried it with .420 as the Time, and got initial V to be -2.465 , can you confirm?

15. Nov 14, 2015

### David112234

so I did it out and got u = 25.6, ,now what?

16. Nov 15, 2015

### PeroK

That number is way too high. How did you calculate that?

17. Nov 15, 2015

### CWatters

That equation comes from the equation of motion..

r = r0 + v0t + 0.5at2

or the SUVAT notation

s = ut + 0.5at2

https://en.wikipedia.org/wiki/Equations_of_motion

It can be applied to any part of the motion on it's own. For example if you apply it to the bit where it goes past the window you can calculate the velocity at the top of the window.