Question

a missile 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 58 feet in the air; after 2 seconds, it is 112 feet in the air. complete the height function, h(x), for this situation. h(x) = enter the correct answer.

Explanation:

Step1: Define quadratic function form

The general form of a quadratic height function for projectile motion (launched from ground, so initial height is 0) is $h(x) = ax^2 + bx$, where $a = \frac{1}{2}g$ (g is gravitational acceleration, $g=-32$ ft/s²), so $a=-16$.

Step2: Use x=1, h=58 to find b

Substitute $x=1$, $h(1)=58$, $a=-16$ into the function:
$58 = -16(1)^2 + b(1)$
$58 = -16 + b$
$b = 58 + 16 = 74$

Step3: Verify with x=2, h=112

Substitute $x=2$, $a=-16$, $b=74$ into the function:
$h(2) = -16(2)^2 + 74(2) = -64 + 148 = 84$
This does not match the given 112, so we use the general quadratic form $h(x)=ax^2+bx+c$. Since launched from ground, $c=0$, so $h(x)=ax^2+bx$. Set up system of equations:
For $x=1$, $h=58$: $a + b = 58$
For $x=2$, $h=112$: $4a + 2b = 112$

Step4: Solve the system of equations

Divide the second equation by 2: $2a + b = 56$
Subtract the first equation from this new equation:
$(2a + b) - (a + b) = 56 - 58$
$a = -2$
Substitute $a=-2$ into $a + b = 58$:
$-2 + b = 58$
$b = 60$

Step5: Confirm the function

Substitute $a=-2$, $b=60$, $c=0$ into $h(x)=ax^2+bx+c$:
$h(x) = -2x^2 + 60x$
Verify with $x=1$: $-2(1)^2 + 60(1) = -2 + 60 = 58$
Verify with $x=2$: $-2(4) + 60(2) = -8 + 120 = 112$

Answer:

$h(x) = -2x^2 + 60x$