Free fall question

1. Oct 25, 2008

msc8127

1. The problem statement, all variables and given/known data
a ball is thrown straight down from the top of a 220 feet building with an initital velocity of -22ft/s. What is the velocity of the ball after 3 seconds? What is the velocity of the ball after falling 108 feet?

2. Relevant equations
s(t) = -16t^2 + Vi(t) + si vi = initial velocity, si = initial position

3. The attempt at a solution
s(t) = -16t^2 +Vi(t) +si
= -16t^2 - 22ft/s + 200ft

V(t) = -32(t) - 22ft/s
V(3) = -96 ft/s - 22ft/s = -118 ft/s

I can seem to figure out how to determine the velocity of the ball after it has dropped 108ft.

I should be able to do this, but i'm missing something, somewhere on what should be pretty basic.

Thanks for any guidance!

2. Oct 25, 2008

nrqed

There are two ways to do it.

The quickest way is to use $$V_f^2 = V_i^2 - 2 g \Delta y$$.

The other ways is to use the equation $$y= V_i t -1/2 g t^2$$ to first find the time to drop 108 feet and then to plug this into $$V_f = V_i - g t$$

3. Oct 25, 2008

msc8127

Thanks for the reply. I have a solution that makes sense now. :)