SOLUTION: Height of a ball. If a ball is thrown straight upward at 40 feet per second from a rooftop 24 feet above the ground, then its height in feet above the ground t seconds after it

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Question 251522: Height of a ball. If a ball is thrown straight upward at
40 feet per second from a rooftop 24 feet above the
ground, then its height in feet above the ground t seconds
after it is thrown is given by
h(t)= 16t2 + 40t + 24.
a) Find h(0), h(1), h(2), and h(3).
b) Rewrite the formula with the polynomial factored
completely.
c) Find h(3) using the result of part (b).
If you can, please show the detailed work, thank you

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Height of a ball. If a ball is thrown straight upward at
40 feet per second from a rooftop 24 feet above the
ground, then its height in feet above the ground t seconds
after it is thrown is given by
h(t)= 16t^2 + 40t + 24.
a) Find
h(0) = 24
h(1) = -16 + 40 + 24 = 48
h(2) = -64 + 80 + 24 = 60
h(3) = -144 + 120 + 24 = 0
------------------------------------
b) Rewrite the formula with the polynomial factored
completely.
h(t) = (-8)(2t^2 - 5t -3)
h(t) = -8[2t^2 -6t + t - 3]
h(t) = -8[2t(t-3)+(t-3)]
h(t) = -8(t-3)(2t+1)
-------------------------------
c) Find h(3) using the result of part (b).
h(3) = -8(0)(7) = 0
===============================
If you can, please show the detailed work, thank you
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Cheers,
Stan H.