Who Threw the Ball First in This Physics Puzzle?

[snowsnowrain]

Homework Statement

A child stands on a balcony directly above a friend 5.00
m below. Each throws a ball toward the other with the
same initial speed of 10.0 m/s. If each catches a ball at
the same instant, which child threw a ball first? How
much later was the second ball thrown?

Homework Equations

I used x = xi + vi(t) + 1/2(a)(t^2)
Where x is the vertical distance.

The Attempt at a Solution

By just thinking about the question, the person below the balcony should throw the ball first if both people catch balls at the same instant.

I'm not sure if vf can equal 0 m/s or not so I used x= xi + vi(t) + 1/2(a)(t^2).
I let the positive x be upward.

Bottom person:

x = +5.00 m
xi = 0
vi = +10 m/s
a = -9.81 m/s^2

-4.905(t^2) + 10(t) - 5

Used the quadratic formula to get t = .87885 sec, 1.159878

Top person:

x = 0.00 m
xi = 5.00 m
vi = -10 m/s
a = +9.81 m/s^2

4.905(t^2) - 10(t) + 5

But doing this makes the t for top person and t for bottom person equal the same.

 

[Nathanael]

Top person:

x = 0.00 m
xi = 5.00 m
vi = -10 m/s
a = +9.81 m/s^2

Why has gravity flipped for the top person?

 

[snowsnowrain]

One of them will be throwing the ball with the help of gravity and I think it should be the top person.

 

[Nathanael]

Gravity points in the same direction regardless of which way the ball is thrown.

Your intuition about the ball being thrown "with the help of gravity" is manifest in the fact that the initial velocity and gravity will have the same direction (both negative). The direction of gravity will not be reversed though.

 

[snowsnowrain]

Thanks a lot. I figured out the top persons time to equal 0.415 sec using the quadratic formula.
I use the bottom time - top time = How much later the second ball was thrown.
If my equation for the bottom person is correct, how am I suppose to know which time value to use? Will it always be the first time (lowest time value)?

 

[Nathanael]

If you imagine the path of the bottom person's ball if the the top person did not catch it: it will go up past the top person, and then come back down past the top person. So the smaller time will be when it first passes the top person (on the way up) and the larger time will be when it again passes the top person (on the way down).

The problem actually didn't specify that the top person catches it on the way up, so in that sense there are multiple solutions. If I had to guess, I would say they want you to assume that it is caught on the way up, but if so then they should've said something about that.

 

[snowsnowrain]

Thanks a lot. You've been very helpful.