# Not sure if this approach is correct for ball falling past a window

1. Jun 20, 2017

### deuce123

1. The problem statement, all variables and given/known data
1. A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person looking through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?

Is it correct to assume that since the average velocity of the ball passing the window is 10 m/s, I can assume it took about 1.22s after the initial position due to gravity to reach the overall average velocity? This is the approach I took to the problem and would appreciate it if anyone can tell me if its correct or not. If not, please explain how too approach the problem. Thank you!

2. Jun 20, 2017

### Orodruin

Staff Emeritus
How did you arrive at this number? How does it answer your question?

Note that the object continues accelerating while falling, do you think this is relevant to your answer? Why/why not?

3. Jun 20, 2017

### Jamison Lahman

No. I believe the average velocity of an object falling with constant acceleration occurs at about 1/3 the distance it travels. The ball spends more time at slower speeds as it is initially accelerating. I will see if I can find the proof.

In the meantime, here is a simple example: the average speed of a ball traveling the distance L/2 at 10m/s and then the distance L/2 at 20m/s is not 15m/s.

4. Jun 20, 2017

### Jamison Lahman

$$t_{slow} = s/v = L/20$$
$$t_{fast} = s/v = L/40$$
Where s is the distance traveled and v is the speed. The average velocity is then:
$$\bar v = \frac{distance}{time} = \frac{L}{\frac{2L}{40}+\frac{L}{40}} = 40/3 \text{m/s}$$
40/3 =/= 15

5. Jun 20, 2017

### deuce123

I used it in turn too calculate the velocity and then ultimately the position. I just divided the gravitational acceleration by the average velocity for the time. I thought about it alittle bit more, so the velocity at the top of the window would be different than the average, so therefore i can't assume that it was dropping for 1.22s beforehand? Also whats another way too approach this problem? Also if anyone can provide me with the correct answer so i can know if i did it correctly would be greatly appreciated.

6. Jun 20, 2017

### deuce123

is it correct too assume for this that tfast covers 2x the ground as tslow? I'm not exactly understanding it well, thank you though.

7. Jun 20, 2017

### Orodruin

Staff Emeritus
You should quote exactly what you did, the formulas you used, and the numbers you put in. Describing it in vague terms does not help us understand what you have actually done.

Assume that it starts a distance $s$ above the window. After how long will it reach the top of the window? After how long will it reach the bottom of the window? What is the difference between these times?

Nobody here will do that until you have solved the problem for yourself. We will give you hints and suggestions and tell you when you have done it correctly, but not solve the problem for you.

8. Jun 20, 2017

### Orodruin

Staff Emeritus
This is false. It depends on the initial velocity. Consider a situation where $v \gg a/t$, where $t$ is the time-scale of the observation.

9. Jun 20, 2017

### Jamison Lahman

I take "dropped" to mean no initial velocity. If the initial velocity is not 0, this problem has infinitely many solutions. The average position of an object in constant acceleration with no initial velocity is 1/3 the distance.

Anyways, I did misinterpret OP's original thinking. We can use the average velocity and the acceleration to find the time it takes to get to that speed, but we don't yet know the position this occurs in regards to the window (and it does not occur at 1m).
I found it easiest to find the velocity as it first enters the view of the window. Do you know how to calculate the change in speed given a distance, acceleration, and time? How can we relate the average velocity to the initial and final velocities of the object while in the window? See if this page helps you: http://www.physicsclassroom.com/class/1dkin/Lesson-6/Kinematic-Equations
I got 4m<distance<5m. That's not the exact answer obviously, but it may help as a reference

Last edited: Jun 20, 2017
10. Jun 21, 2017

### haruspex

@deuce123: Please follow the line Orodruin is taking you in post #7. You can ignore the rest of this post.

@Jamison Lahman: Not that it helps at all in solving the question, but
No. It's a quarter.
$\Delta s = \frac 12 a\Delta t^2$
$v_{avg}=a\frac 12\Delta t=v|_{\frac 12\Delta t}$
So average velocity is reached at $t_{v_{avg}}=\frac 12\Delta t$
$s|_{v_{avg}}=\frac 12 a(\frac 12\Delta t)^2=\frac 18 a\Delta t^2=\frac 14\Delta s$

Yes.

11. Jun 21, 2017

### Orodruin

Staff Emeritus
Yes, but the window of observation (haha :p) is not from the drop time and so your idea is not applicable to the average speed that you can observe here.