# Two balls meet in mid air

1. Oct 2, 2012

### steven1495

1. The problem statement, all variables and given/known data
Two balls are juggled so that each ball is in the air for 2.000s, and the second ball is thrown UP at the instant the first starts down. Find the height above the throwing hand where the two balls pass.
(Assume gEARTH = 9.800m/s2 down)

TT=2.000s
g=9.800m/s2 down

2. Relevant equations
d=vft-(1/2)at2
d=(vf+vi)/2

3. The attempt at a solution

Finding the max height
d=vft-(1/2)at2
d=0(1.000)-(1/2)(9.800)(1.000)2
d=4.900m

Finding the Initial Velocity
d=(vf+vi)/2
4.900=(0+vi)/2
Vi=9.800m/s

This is all I could get. Many thanks to whoever helps steer me in the right direction

2. Oct 2, 2012

### azizlwl

Let start the time when ball A starts to descent and ball B start to ascent is zero.
Let at time t they passed each other at height equal to h.

Or you can derive 2 equations of height ball A and B and solve for H.

Last edited: Oct 2, 2012
3. Oct 2, 2012

### steven1495

Can you elaborate on this a bit?

4. Oct 2, 2012

### azizlwl

Just write whatever equations of height for ball A and ball B that you might think of and we will make correction if they wrong.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook