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March 6, 2015

March 6, 2015

Posted by **Rachel** on Thursday, January 31, 2013 at 1:13am.

- calculus 2 -
**Reiny**, Thursday, January 31, 2013 at 8:46amshould have been g = -32

then

v = -32t + c, but when t = 0 , v = 0 , so c = 0

we have v(t) =-32t

then

a(t) = -16t^2 + k, when t = 0 , s = 400 , so k = 400

so our distance equation is

s(t) = -16t^2 + 400

when it hits the ground, s = 0

0 = -16t^2 + 400

16t^2= 400

t^2 = 25

t = 5

which was given, and is correct

if we didn't know the acceleration, let it be a

our distance equation would have been

s(t) = (a/2)t^2 + 400

at half the time, t = 2.5

0 = (a/2)(6.25) + 400

a/2 = -400/6.25

a = -800/6.25 = -128 ft/sec^2

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