# Problem about Free Falling Objects

1. Oct 19, 2012

### Ammar w

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

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

2. Relevant equations

yf = yi + vit + ½ at2

3. The attempt at a solution

The right solution :

The ball is thrown 2 sec before the rock
yball = ½ at2 + vit + yi
yball = ½(-10)(t+2)2 + 20(t+2)
yball = -5t2 + 20

yrock = ½ at2+ vit + yi
yrock = ½(-10)t2 + 24t
yrock = -5t2 + 24t

yball = yrock (they pass each other)
-5t2 + 20 = -5t2 + 24t
t = 5/6 seconds

yrock = -5(5/6)2 + 24(5/6)
yrock = 17 meters

but I don't know why he considered (t) of the ball (t+2) althought the ball is thrown 2 seconds before the stone???
isn't it true if :
t of the ball = t-2
t of the stone = t
?

Last edited: Oct 19, 2012
2. Oct 19, 2012

### frogjg2003

Where is the ball at t=0 and where is the rock at t=0? The reason you use t+2 instead of t-2 is because the equation you used for the rock assumed it was y=0 at t=0, but it was at y=0 at t=-2. t-(-2)=t+2

3. Oct 19, 2012

### Ammar w

Thank you. that helps.
but I will be pleased with more clarification.

4. Oct 19, 2012

### frogjg2003

The equation you used both times, $y=v_0t-\frac{1}{2}gt^2$ is for a projectile starting at position 0m. You have to have t the same for both objects, and the ball is not at its starting position of 0m at t=0s, but it was at t=-2s.