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October 1, 2014

October 1, 2014

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

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 9:00amYour 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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