# MATH

The height h (in meters) of a cannonball t seconds after it is fired into the air is given by
h=-4t^2+16t+9.

a) What is the initial height of the cannonball? What feature does this correspond to on the graph?

b) Find the height of the cannonball after 1 second.

c) What is the maximum height of the cannonball? When does it reach the maximum height?

d)How long does it take for the cannonball to fall back to earth?

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1. a) Hey, you know what h is when t = zero!

b ) h = -4(1) + 16(1) + 9
= 21

c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola
t^2 - 4 t = -h/4 + 9/4

t^2 - 4 t + 4 = -h/4 + 9/4 + 16/4

(t - 2)^2 = -(1/4) (h - 25)
so
max height = 25 at t = 2

d ) 0 =-4t^2+16t+9
or
t^2 - 4 t - 9/4 = 0

t = [ 4 +/- sqrt(16 -81/4) ] /2
no real solution
The reason is that your original equation is in error
h=-4t^2+16t+9.
It should be assuming feet and seconds and g = -32 ft/s^2
h = (1/2) g t^2 + Vi t + Hi
h = -16 t^2 + Vi t + Hi
I suspect a typo

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2. I agree with Damon that your gravitational constant is off.
Since you are using metric , your equation should be

h = -4.905t^2 + 16t + 9
let's ballpark it to

h = -5t^2 + 16 + 9

a) .....

b) sub in t = 1 , h = -5+16+9 = 20 m

c) for t of the vertex ...
t =-b/(2a) = -16/-10 = 1.6
h = -5(1.6^2) + 16(1.6) = 9 = 21.8

Maximimum height of 21.8 m after 1.6 seconds

d) -5t^2 + 16t + 9 = 0
5t^2 - 16t - 9 = 0
t = (16 ± √436)/10
= 3.69 or a negative

it will hit the ground after 3.69 seconds

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3. d) use quadratic formula
-b +- square root of b^2 - 4 times a times c devided by 2a
= -16+-square root 400 devided by -8
= 4.5 secounds.

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