Question

a rock is launched from a cannon. its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. after 1 second, the rock is 148 feet in the air; after 2 seconds, it is 272 feet in the air. complete the height function, h(x), for this situation. h(x) = enter the correct answer.

Explanation:

Step1: Define quadratic form

The general height function for projectile motion (in feet) is $h(x) = -16x^2 + v_0x + h_0$, where $v_0$ is initial velocity, $h_0$ is initial height. We also know:
When $x=1$, $h(1)=148$: $148 = -16(1)^2 + v_0(1) + h_0$ → $v_0 + h_0 = 164$
When $x=2$, $h(2)=272$: $272 = -16(2)^2 + v_0(2) + h_0$ → $2v_0 + h_0 = 336$

Step2: Solve for $v_0$

Subtract first equation from second:
$(2v_0 + h_0) - (v_0 + h_0) = 336 - 164$
$v_0 = 172$

Step3: Solve for $h_0$

Substitute $v_0=172$ into $v_0 + h_0 = 164$:
$172 + h_0 = 164$
$h_0 = 164 - 172 = -8$

Step4: Build height function

Substitute $v_0=172$, $h_0=-8$ into the general form:
$h(x) = -16x^2 + 172x - 8$

Answer:

$h(x) = -16x^2 + 172x - 8$