# Throwing and dropping two balls: Find height where they meet

1. Aug 21, 2015

### Tulpje

1. The problem statement, all variables and given/known data
You throw one ball vertically (V0= 15m/s) on top of a 100m tower. Two seconds later you release another ball from rest from the same tower.
Create a motion equation for both balls
At what moment do they pass each other?
At what height?

2. Relevant equations
s=vt+1/2at^2

3. The attempt at a solution
First I thought I could figure out how long it took the first stone to come back down to the height of the top of the tower (where the other ball was released) and came up with about 3.06s via
V0/a or 15/9.8 then multiply by two. Because this would reach 0 one second after ball two, I then made s=s with t-1 for ball one and t for ball two.

I then came out to a formula: -4.9t^2=15t-15-4.9t^2+9.8t-4.9
After solving I got t=0.407 but if I insert this into the formulas (1/2*-9.8*t^2) and 15*(t-1)+1/2*-9.8*(t-1)^2 to find the distance that they have traveled it never works out.
I have tried this in many different ways, including starting from the beginning with t-2 for ball two and still I come up with numbers that don't work.

I think I am having trouble visualizing the time that the balls are actually in the air.
Am i wrong in first trying to figure out when the first ball reaches s=0 (top of tower?) Is there a way that I can do this including the entire journey of the first ball in the equation?
If I use t-1 what exactly does this mean? That the ball is in the air for one second less?

If anyone could help I would be very appreciative. I am self studying physics for an exam, and I have a lot of trouble understanding it, so I really need to see clear steps to follow. Thank you!

2. Aug 21, 2015

### RUber

It sounds like you have a great start. Don't forget the height of the tower.
*edited to correct initial velocity*
Ball 1 has position $s_1 = 100+15t - 4.9t^2$.
Ball 2 has position $s_2 = 100-4.9(t-2)^2$ for t >2.
Total distance travelled for ball 2 is d= 100-s, where s is where they meet. For ball 1 it is (max height-100)x2 + d.

3. Aug 21, 2015

### Tulpje

Oh :/ I cant believe I forgot that! Is the equation then actually:
s=s0+v0t+1/2at2 ?

I got t=4.26s and they reach each other at about 75m.

Thank you!!
Why does it not work if I do it the first way? If I determine how long it takes for ball one to be thrown up and come back down to the height of the tower and I use appropriate values for t or t-1 shouldn't it also work?

4. Aug 21, 2015

### RUber

This doesn't look quite right. First, rounding 3.06 seconds to 3 might influence your solution depending on how many significant digits you want to keep. Second, if you are only looking at the downward part of the problem, your initial velocity of the first ball is pointing down.
You should have had $-15(t-1) - 4.9(t-1)^2 = -4.9 t^2$.
Since you started with the wrong velocity direction, the solution you found was for t<1, or before your first ball returned to the top of the tower.

Graphically, here is what the two position functions should look like, along with an approximate reference point.

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5. Aug 21, 2015

### Tulpje

Thank you! It makes sense now

6. Aug 21, 2015

### PeroK

What about considering the relative motion between the balls?