# Gravity and Kinematics

1. Sep 11, 2008

A ball is dropped from rest from the top of a building and strikes the ground with a speed Vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is Vo=Vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning.

Hmm...
I thought of using kinematics to find if both times would be equal to each other when they arrive to halfway x = vt +1/2at^2, but the only difference between the balls would be their direction considering the velocities are similar.
What to do? Ty

2. Sep 12, 2008

### alphysicist

Re: Kinematics

You can definitely use kinematics to find the answer.

But conceptually, think about the fact that one starts out moving slow and speed up, while the other starts out moving fast and slow down. How does that affect the spot where they pass each other?

3. Sep 12, 2008

Re: Kinematics

Thanks!

So:

Throwing up: Starts fast and slows down
Dropping: Starts slow and speeds up
In this sense, greater velocity equals greater distance traveled. Therefore the balls should cross each other above the halfway point.

Is that true?

4. Sep 12, 2008

### alphysicist

Re: Kinematics

That sounds right to me.

5. Sep 12, 2008

### Johannes

Re: Kinematics

however, intuition often plays tricks on us, so, just for practice, you should consider doing the maths behind that and find out where exactly they meet, now since you know the fundamental difference between both motions.

6. Sep 12, 2008

Re: Kinematics

Thanks!

@above: We are provided with only one value such that a = 9.8m/s^2. Thus, how can we calculate it's location?

7. Sep 12, 2008

### alphysicist

Re: Kinematics

You can calculate at what fraction of the height of the building they cross (the halfway mark, 1/10 of the way up, etc.).

8. Sep 12, 2008

Re: Kinematics

to think about it, i'm not sure how to prove that the balls would arrive at above the halfway point. any suggestions? possibly using equations?

9. Sep 12, 2008

### alphysicist

Re: Kinematics

To use equations, just set the height building to be $h$ (or choose a numerical height) and solve the two motions for the time in which they are at the same position (calculating some needed quantities like the final speed of the dropped ball along the way). Once you find that position, you can compare it to the building height to answer the question.

10. Sep 14, 2008

Re: Kinematics

Hmm, but we don't know the height of the building? :S

11. Sep 14, 2008

### alphysicist

Re: Kinematics

What I meant was to set the building height to be an unknown variable $h$, and then solve for the height at which the balls cross.

Then, just as an example, if you found the balls cross at the vertical position $y=\frac{1}{4}h$ then you know that they cross below the midpoint of the building.

12. Sep 14, 2008

Re: Kinematics

Ight! Thank you! :)

13. Sep 14, 2008

### alphysicist

Re: Kinematics