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?

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

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

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.

a) To find the initial height of the cannonball, we need to determine the height when t = 0. In the given equation, substitute t = 0:

h = -4(0)^2 + 16(0) + 9
h = 0 + 0 + 9
h = 9

Therefore, the initial height of the cannonball is 9 meters. This corresponds to the y-intercept of the graph, where the cannonball is at ground level and has not yet been fired.

b) To find the height of the cannonball after 1 second, substitute t = 1 into the given equation:

h = -4(1)^2 + 16(1) + 9
h = -4 + 16 + 9
h = 21

So, the height of the cannonball after 1 second is 21 meters.

c) To find the maximum height of the cannonball, we need to determine the vertex of the quadratic equation. The vertex form of a quadratic equation is given by:

h = a(t - h)^2 + k

Here, a = -4, h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.

The formula for the x-coordinate (h) of the vertex is given by:

h = -b/2a

In our case, a = -4 and b = 16. Substituting these values:

h = -16 / (2 * -4)
h = -16 / -8
h = 2

The x-coordinate of the vertex is 2 seconds. To find the y-coordinate (k) of the vertex, substitute t = 2 into the original equation:

h = -4(2)^2 + 16(2) + 9
h = -4(4) + 32 + 9
h = -16 + 32 + 9
h = 25

So, the maximum height of the cannonball is 25 meters, and it is reached after 2 seconds.

d) To find how long it takes for the cannonball to fall back to earth, we need to determine the time when the height (h) is equal to zero.

Setting h = 0 in the original equation:

0 = -4t^2 + 16t + 9

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -4, b = 16, and c = 9. Substituting the values:

t = (-16 ± √(16^2 - 4 * -4 * 9)) / (2 * -4)
t = (-16 ± √(256 + 144)) / -8
t = (-16 ± √(400)) / -8
t = (-16 ± 20) / -8

Now, calculate both possibilities:

t1 = (-16 + 20) / -8
t1 = 4 / -8
t1 = -0.5

t2 = (-16 - 20) / -8
t2 = -36 / -8
t2 = 4.5

The cannonball falls back to earth after approximately 0.5 seconds (t = -0.5) and 4.5 seconds (t = 4.5). As we consider time in a real-world context, the negative value is not meaningful, so we can disregard it. Therefore, it takes approximately 4.5 seconds for the cannonball to fall back to earth.