# Homework Help: Kinematics simultaneous motion problem!

1. Sep 26, 2006

### physstudent1

"You are on top of a building 186 meters high you throw a ball downard at a speed of 18m/sec"

After calculating time for hitting the ground final velocity and average velocity the next question says:

"Suppose instead that at the same moment that this ball was thrown downward from the roof a second ball was thrown upward from the ground with a speed of 32m/s"

I know that when theres simultaneous motion you are supposed to find what the two objects have in common but I can't seem to find it.

I found the time to hit the ground for the first ball was 4.59, its final velocity was -62.98m/s and its average velocity was -40.52m/s.

I figured the time it takes the ball from the ground to reach a velocity of 0 is 3.27seconds and the height it reaches is 52.32 meters I'm not sure what to do now though, somebody please give me a hint or steer me in the right direction. Thank you.

2. Sep 26, 2006

### skeeter

for the ball thrown downward ...
h = 186 - 18t - 4.9t^2

for the ball thrown upward ...
h = 32t - 4.9t^2

they meet when h = h ...

186 - 18t - 4.9t^2 = 32t - 4.9t^2

solve for t (very easy), then find h.

3. Sep 26, 2006

### physstudent1

thanks..?

Okay, I understand that they both will hit when h = h, but where are you getting those two equations from, and I calculated a time of 3.72 seconds for them to collide, and a height of .85 meters please let me know if this is right.

Can anyone let me know if these answers are correct?

Last edited: Sep 26, 2006