Posted by Rachel on Thursday, January 31, 2013 at 1:13am.
should have been g = -32
v = -32t + c, but when t = 0 , v = 0 , so c = 0
we have v(t) =-32t
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
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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