# Free fall problem

1. Jun 19, 2014

[EDITED. Thanks to Doc Al, Tanya Sharma and dauto]

1. The problem statement, all variables and given/known data

A ball is dropped from rest from the top of a building of
height h. At the same instant, a second ball is projected vertically
upward from ground level, such that it has zero speed when it
reaches the top of the building.

(A) At the time when the two balls pass each other -
which ball has the greater speed, or do they have the same speed?

(B) Where will the two balls be when they are alongside each
other: at height h/2 above the ground, below this height, or above
this height?

2. Relevant equations

Equations of 1D motion for a constant acceleration of -g

3. The attempt at a solution

First here is the diagram I drew:

Now for part (A):
I suppose that the second ball was thrown up with velocity v0
And the balls pass each other at time t
So for the first ball v = -gt
And for the second ball v = v0 -gt
So speed of the first ball |-gt|
Speed of the second ball |v0 -gt|
Now which one is greater? How can I tell??

I shall post my attempt for second part later but someone please help me deal with this.

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Last edited: Jun 19, 2014
2. Jun 19, 2014

### Staff: Mentor

You cannot ignore all signs. Do you think that the ball thrown upward increases its speed as it rises?

3. Jun 19, 2014

Sorry. I meant to say that the second ball was thrown up with velocity v0

Last edited: Jun 19, 2014
4. Jun 19, 2014

### Tanya Sharma

Since upward is positive shouldn't the initial velocity v0 be positive ?

5. Jun 19, 2014

Of course! (Hey what's wrong with me!)
Thanks.

6. Jun 19, 2014

### dauto

Why is the initial speed of the second ball negative? are you sure about that?

7. Jun 19, 2014

That was a silly mistake. Thanks for pointing out.

8. Jun 19, 2014

### Bandersnatch

The first part asks you about the speed at the time when they pass each other. Why don't you start by finding this time? Which equation you ought to use?(hint: they are passing each other)

9. Jun 19, 2014

Ok. i think when the balls pass each other they have same position (same y).
For the first ball y = -0.5gt2
And for the second ball, the initial position of the ball is -h and if final position is y and initial velocity vo then y - (-h) = v0t - 0.5gt2
Finally for the second ball y = v0t - 0.5gt2 - h

Equating this two, -0.5gt2 = v0t - 0.5gt2 - h which gives
t = h/v0

Is everything ok?

10. Jun 19, 2014

### Bandersnatch

Righty-o.

Now, why won't you use it to find both of the velocities.

11. Jun 19, 2014

For ball 1, v = -gt = -gh/v0
For ball 2, v = v0 - gt = v0 - gh/v0

But now which one is greater?

12. Jun 19, 2014

### Bandersnatch

Good question! These equations by themselves don't help us much.

Luckily, we can try to look for additional clues somewhere else. Can you try and write the conservation of energy for either of the balls?

13. Jun 19, 2014

Hey, maybe I can do something else too. (I was thinking in this line for a while)
We know, v2 = v02 - 2gy
But for ball 2 when it moves up h height, v = 0 (as given in question) and displacement (0 - (-h)) = h
So then, at h, for ball 2, 0 = v02 - 2gh
or v0 = $\sqrt{2gh}$

We know t = h/v0 = h/$\sqrt{2gh}$
So for ball 1, v = -gt = -g$\sqrt{h/2g}$ = -$\sqrt{gh/2}$ = -v0/2
For ball 2, v = v0 - gt = v0 - v0/2 = v0/2

So when they meet their speed is same.... What do you think? Is this ok?

14. Jun 19, 2014

### Staff: Mentor

Hint: h and v0 are not independent. (Express v0 in terms of h.)

(Edit: Looks like you figured that out!)

15. Jun 19, 2014

### Bandersnatch

Adjoint, that is exactly right. You actually did use the conservation of energy, only rather than write it out, you used the ready-made kinematic equation. Otherwise it's all the same.

Had you written the conservation of energy it'd look like this:
$E_{p_0}+E_{k_0}=E_{p_1}+E_{k_1}$ where 0 is inital, 1 is final
for ball 1: $0+0=-mgh+\frac{1}{2}mV_1^2$ so we end up with $\frac{1}{2}mV_1^2=mgh$ and extracting V:
$V_1=\sqrt{2gh}$
for ball 2 it's of course symmetrical, as we know it has to end up with the same potential energy as the first one, so it's initial kinetic energy must be the same as ball 1's final one. This lets us identify V1 as being equal to V0
$V_0=\sqrt{2gh}$

16. Jun 19, 2014

Okey, so then I hope answer of part (A) is done.

Now, for part (B), which asks: Where will the two balls be when they are alongside each other?

We already know t = $\sqrt{h/2g}$ is the time when the balls meet.
So, we can take either ball 1 or ball 2 and find its position at time t = $\sqrt{h/2g}$ And that will be the answer of (B), correct?

(for ball 1) y = -0.5gt2 = -0.5g($\sqrt{h/2g}$)2 = - h/4

That's the answer of B, right?

17. Jun 19, 2014

### Bandersnatch

Looks like you've got it!

Intuitivelly, it's not hard to imagine too. The upward ball will cover the first half of the way super fast, going ever slower on the second leg of its flight. The downward ball does the opposite. It kinda makes sense for them to meet way above the halfway point.

18. Jun 19, 2014

### Staff: Mentor

Yes, but express it to answer the question exactly as asked. (Since they express heights above the ground, so should you.)

19. Jun 19, 2014

### Orodruin

Staff Emeritus
You are correct, so since you have solved the problem let me try to offer a view on how to solve it without solving a single equation.

A) Since both balls are at rest at y=0, they have the same total energy (really the same total energy/mass, but this is a minor detail, let us work with unit masses). When they are at the same height they have the same potential energy and therefore the same kinetic energy, hence same speed.

B) For each time before they meet, the top ball has a lower velocity, since it has a larger potential energy (and therefore lower kinetic energy). Thus, when they meet, the ball coming from below will have traveled further. The balls thus meet above the mid-point.

20. Jun 19, 2014