Posted by Audrey on Friday, February 3, 2012 at 7:46am.
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!

Calculus  Reiny, Friday, February 3, 2012 at 9:00am
Your equation
s(t) = 16t^2  22t + 220 is correct and tells you
how high the object is at any time t
But they are asking for the velocity after it has fallen 108 feet, not when it is 108 feet high
It started at 220, so after falling 108 feet, the height will be 112
so 112 = 16t^2  22t + 220
16t^2 + 22t  108 = 0
t = (22 ± √7396)/32
= 2 or a negative, (ahhh, it would have factored)
so take v(2) , that should give you the answer you need
check:
when t=0 , s(0) = 220  it has fallen 0 feet
when t=1 , s(1) = 182  it has fallen 38 ft
when t=2 , s(2) = 112  it has fallen 108 ft
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