# Three balls are dropped, find position of each ball

1. Sep 10, 2012

### Phys121VIU

1. The problem statement, all variables and given/known data
Ball A is dropped at t=0 from a window. At t=6, the ball hits the ground. Neglect air resistance for question

a) Suppose that at t=1, Balls B and C are thrown vertically down from the same window. Ball C has a greater initial velocity than Ball B, as shown by their positions at t=2 in diagram. On the diagram show the positions of Balls B and C at times 3,4,5...unit they hit the ground

b)At what times, approx, do the balls hit the ground?

c)Does Ball B pass A? if so at what time? Before hitting the ground so Balls A and B have the same velocity? at what times?

d) Does Ball C pass Ball A? if so at what time? Before hitting the ground so Balls A and B have the same velocity? at what times?

3. The attempt at a solution

I have attached a picture of the diagram provided. What im confused about is in part a) how to find out the location of Ball B and C at time=3,4,5.. To me it seems there is insufficient data provided..please help me on the right track, but no answers please!

#### Attached Files:

• ###### Photo 2012-09-09 10 37 15 PM.jpg
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2. Sep 10, 2012

### ehild

The diagram shows the position of the balls after 1 s fall. From the position of ball A, you can find the length of scale. From the positions of the other balls, you can find their initial velocity. Knowing the initial velocity, you can determine the position of a falling ball.

ehild

3. Sep 10, 2012

### PeterO

Unfortunately, the position of object A indicates a lot of air resistance acting.

For an object dropped where air resistance can be ignored, it falls 3 times as far in the second time interval [second?], compared to how far it falls in the first time interval [second]. It then goes on to fall 5 times as far in the third time interval compared to the first.
The data you are given for body A it falls only twice as far in the second time interval, and 3 times as far in the 3 rd time interval.

That makes the question pretty well un-answerable, though there is probably the implication that object B never catches object A while object C has some chance of catching up.

example: distance fallen when g = 10 ms-2.
1st second 5m [an object drops 5m in one second]
2nd second 15m [an object falls 20m in 2 seconds]
3rd second 25m [an object falls 45m in 3 seconds]

4. Sep 10, 2012

### ehild

You are right, the figure is rather mysterious. the vertical axis can not show the positions at 1, 2, ..6 s of ball A if it is dropped and there is no air resistance.

ehild

5. Sep 12, 2012

### Phys121VIU

why not?

6. Sep 12, 2012

### PeterO

because the distances covered in each time unit are not appropriate.

The positions shown in your figure are :

0, 1, 3, 6, 10, 15, 21 meaning distances travelled in each time interval are 1,2,3,4,5,6

For an object falling under gravity, they should be:

0,1,4,9,16,25,36 meaning distances travelled in each time interval are 1,3,5,7,9,11

remember the formula

s = ut + 1/2at2 : it is easy to see why the position increases as the perfect square values.

7. Sep 12, 2012

### PeterO

btw - this is what I think the problem was trying to say:

The first second of B's motion is the same length as second second of A's motion.
This implies that while A was dropped, B was thrown 1 second later at a speed exactly that which A had gained in its first second of falling.

For example
If you drop a stone on Earth, it reaches 9.8 m/s after falling for 1 second.
You could then throw B down at 9.8 m/s.

Since both stones have the same speed at time t = 1, and are under the influence of the same gravity, so will have the same acceleration.
That means they will, at all times, have the same speed.
Thus B will begin 4.9 m behind, and remain 4.9 m behind - it never catches up, so never passes.

Now to C.
In its first second it traveled further in its first second than B, so it began faster than B [and faster than the speed A had reached].
Due to all three being under the influence of the same gravity field, C will at all times be travelling faster than the other two. It will thus be catching up to A.
If you had correct information [ie positions outlined in my previous post] you would be able to work out whether C catches/passes A before they hit the ground.

Once again to approximate and make the best of a bad problem:

The distance travelled by C in its 1st second is shown as 2.5 squares. A is shown as 2 squares in its 2nd second and 3 squares in its 3rd second.
One implication is that C was thrown at a speed equal to what A would have reached after 1.5 seconds.

On Earth that would mean C is thrown at 14.7 m/s. It is gaining on A at 4.9 m/s [14.7 - 9.8] but beginning only 4.9 m behind, so it will catch up at time t = 2 and easily win the race to the ground.

Since the Graph supplied is so flawed, it is difficult to establish how far it actually is to the ground so difficult to see by how much C wins the "race".

8. Sep 12, 2012

### PeterO

Finally what I expect was expected of your answer.

You would draw B as always 1 square behind A. ie it is shown at t=1 at zero, t = 2 at two, so t=3, 3 squares further down, t = 4 4 squares further down etc

It is probably expected that where C is shown at zero and 2.5, the next dots would be 3.5 further [ie at square 6] 4.5 further down [ie at 10.5] then 5.5 further [at 17] etc.

That would also match with the idea of overtaking at t= 3

[I said C overtook at time t = 2 in previous post, I meant after 2 seconds of C's motion (which began at t=1) so it should have over taken at t = 3.]

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