Amissile is launched from the ground. its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. after 1 second, the missile is 130 feet in the air; after 2 seconds, it is 240 feet in the air. find the height, in feet, of the missile after 11 seconds in the air.

h(11) = 330 ft

Step-by-step explanation:

We have 2 points with which to find the equation of the projectile.  The standard form for the quadratic, which is the function that models parabolic motion, is

The problem tells us that the missile was launched from the ground, so c (the initial height of the missile) is 0, so we can literally disregard the c in the standard form and use the 2 points to solve for a and b:

At the point (1, 130) we fill in the standard form to get:

which gives us **130 = a + b**.  Now on to the second point, (2, 240):

which gives us **240 = 4a + 2b**

Solve the first **equation for a:

a = 130 - b

and sub that into the second **equation:

240 = 4(130 - b) + 2b and simplify to

140 = b

Now that we know b, we can sub it in to solve for a:

a = 130 - b so a = 130 - 140 so a = -10.  Therefore, the equation of the projectile's motion is

(If you knew your Physics, you'd know that -10x^2 is replacing the pull of gravity in the metric system which is -9.8x^2; that's how we know that this is the correct equation.)

Subbing in 11 for x:

gives us that h(11) = 330 feet

10

step-by-step explanation:

7+3=10