# Kinematics (a little more difficult then I am used to)

1. Jan 23, 2008

### Roger Wilco

1. The problem statement, all variables and given/known data

Ball A is released from rest at height of 40 ft at the same time Ball B is thrown upward from 5 ft off the ground.

If the Balls pass one another at 20 ft above the ground, determine the speed at which ball B was thrown upward.

2. Relevant equations
$v=v_o+at$
$v^2=v_i^2+2a(y-y_i)$
$y-y_i=v_it+.5at^2$

3. The attempt at a solution

I have used the information about ball A to determine the time it takes A to reach a height of 20 ft:
Choosing down as positive, $20=16.1t^2$

thus, t=1.115 s. Now this must be the same time that B reaches 20 ft. I know that g is pulling down on it and it must cover a vertical distance of 15 ft in 1.114 s.

I am just getting stuck in trying to write that last sentence algebraically.

~RW

2. Jan 23, 2008

### Midy1420

You're on the right track. You are looking for the initial velocity of ball B. We know the displacement (15ft), we know the acceleration (-32.2 ft/(s^2)) and you found the time it takes to travel this distance (1.115 s) it takes to reach that value. Which equation would you chose, to solve for Vinitial?

3. Jan 23, 2008

### Roger Wilco

Oh good grief. I feel like such a mutton-head (is mutton-head hyphenated?).

Thanks Midy,
~RW