# Velocity Time Graphs and Questions

1. Oct 16, 2008

### Run4Fun

I've been working on this one problem for a while and I don't really seem to understand it. If you could help me understand how to do this, I would be much obliged.

Two students are on a balcony 19.6 m about the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The socond ball just misses the balcony on the way down.

a. What is the difference in the time the balls spend in the air?

b. What is the velocity of each ball as it strikes the ground?

c. How far apart ar the balls 0.800 s after they are thrown?

I attempted the first one and got it right (by a miracle in my opinion)- 3.0 Seconds

But I seem to be stuck when it comes to b.

2. Oct 16, 2008

### MATdaveLACK

The velocities of each ball right before they hit the ground will be equal (since when the ball thrown vertically upward comes back down to the point from which it was thrown, it's velocity will be the negative of it's initial velocity, 14.7 m/s. And that is the same velocity the first ball was thrown straight down with)
You can use kinematic equations to solve for final velocity (displacement = -19.6m, acceleration = -9.8m/s/s, Vi= +/- 14.7 m/s, then solve for Vf)

Last edited: Oct 16, 2008
3. Oct 16, 2008

### LowlyPion

Welcome to PF.

It shouldn't take a miracle to figure the first one.

Whatever the speed thrown upward will be the same speed as it passes the same height on the way down. That simply makes it a calculation of how long to go up from the balcony and back down.
14.7m/s/9.8m/s2 = 1.5 is how long to go up and double it for the up and down. 3 seconds.

For b) since they both pass the balcony at the same speed they will strike the ground with the same speed.
V2 = Vi2 + 2*g*x

For c) you need to use x = Xi + Vi*t +1/2*g*t2 (appropriately set up for each ball) and calculate the positions of each at .8s.