A model rocket is shot up from the roof of a school. The height at any time is modeled by the equation, H = -3t² + 12t + 36, where H is the height in metres and t is the time in seconds.

a) When does the rocket hit the ground?
b) How long does it take for the rocket to pass a window 21 metres above the ground?
c) What is the maximum height of the rocket?

a) To find when the rocket hits the ground, we need to find the time when the height (H) is equal to zero. In other words, we need to solve the equation -3t² + 12t + 36 = 0.

To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b² - 4ac)) / (2a), where a, b, and c are the coefficients from the equation.

In this case, a = -3, b = 12, and c = 36. Plugging these values into the quadratic formula, we have t = (-(12) ± √((12)² - 4(-3)(36))) / (2(-3)).

Simplifying further, t = (-12 ± √(144 + 432)) / (-6).

Taking the square root of 576, we get t = (-12 ± 24) / (-6).

Solving for both values, t = (-12 + 24) / (-6) = 2 and t = (-12 - 24) / (-6) = 6.

Therefore, the rocket hits the ground at 2 seconds and 6 seconds after it was launched.

b) To find how long it takes for the rocket to pass a window 21 meters above the ground, we need to find the time when the height (H) is equal to 21. In other words, we need to solve the equation -3t² + 12t + 36 = 21.

Rearranging the equation, we have -3t² + 12t + 36 - 21 = 0.

Simplifying further, -3t² + 12t + 15 = 0.

Now we can solve this quadratic equation using the quadratic formula:

t = (-(12) ± √((12)² - 4(-3)(15))) / (2(-3)).

Simplifying further, t = (-12 ± √(144 + 180)) / (-6).

Taking the square root of 324, we get t = (-12 ± 18) / (-6).

Solving for both values, t = (-12 + 18) / (-6) = -1 and t = (-12 - 18) / (-6) = 5.

Therefore, it takes -1 seconds and 5 seconds for the rocket to pass a window 21 meters above the ground. Since time cannot be negative in this context, the rocket passes the window at 5 seconds after it was launched.

c) To find the maximum height of the rocket, we need to find the vertex of the quadratic equation. The vertex corresponds to the highest point on the rocket's trajectory.

The vertex of a quadratic equation in the form of y = ax² + bx + c is given by the x-coordinate, which is given by the formula x = -b / (2a).

In this case, a = -3 and b = 12. Plugging these values into the formula, we have x = -12 / (2(-3)).

Simplifying further, x = -12 / (-6) = 2.

Now, to find the corresponding y-coordinate (which represents the maximum height of the rocket), we can substitute this value back into the equation H = -3t² + 12t + 36.

H = -3(2)² + 12(2) + 36.
H = -12 + 24 + 36.
H = 48.

Therefore, the maximum height of the rocket is 48 meters.

a) when it hits the ground -3t² + 12t + 36 = 0

t^2 - 4t - 12 = 0
(t - 6)(t + 2) = 0
take over

b) -3t² + 12t + 36 = 21
t^2 - 4t - 5 = 0
this one also factors nicely, so keep grinding it out

c) H = -3t² + 12t + 36 is a parabola opening downwards, find the vertex and you got it