Question 61·200 Super-Hard SAT Math Questions·Advanced Math

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A quadratic function $h$ models the height, in feet, of a model rocket above the ground $t$ seconds after it is launched.

During its flight (before it hits the ground), the rocket reaches a maximum height of 180 feet. One second after it is launched, the rocket is 160 feet above the ground. The rocket hits the ground 9 seconds after it is launched.

Which choice is the height, in feet, of the rocket above the ground at the moment it was launched?

A rocket’s greatest height is the vertex, the peak of its quadratic height graph, so start with vertex form even when the peak time is unknown. Compare the given heights by how far each is below that peak. Use a Desmos regression to find the peak time, then evaluate the height at launch. The trap is treating the height change as constant each second.

Hints

Hint 1
A maximum of $180$ fits the form $h(t)=180-k(t-v)^2$ , where $v$ is the peak time and $k>0$ . How far below the maximum is the rocket at each of the two times you know?
Hint 2
At $t=1$ , the rocket is $20$ feet below its peak; at $t=9$ , it is $180$ feet below. Each drop is $k$ times a squared distance from the peak time. What cancels when you divide the drops?
Hint 3
The maximum occurs during the flight, so its time cannot be before launch or after landing. Once you find that time, use a given height to find $k$ , then evaluate $h(0)$ rather than $h(1)$ .

Step-by-step

Build the height model from its peak
Step 1
Write the peak in vertex form
Let $v$ be the time of the vertex, the peak, and let $k>0$ . A squared distance is $0$ at $t=v$ and positive at other times, so subtracting it makes $180$ the maximum:
$h(t)=180-k(t-v)^2.$
Step 2
Find the drop at one second
The height at $t=1$ is $160$ feet, so substitute that time and height:
$160=180-k(1-v)^2.$
Rearrange to put the drop alone:
$k(1-v)^2=180-160.$
Subtract:
$k(1-v)^2=20.$
Step 3
Find the drop at landing
Hitting the ground at $t=9$ means the height is $0$ , so:
$0=180-k(9-v)^2.$
Rearrange to put the drop alone:
$k(9-v)^2=180.$
Step 4
Compare the two drops
Below a quadratic maximum, the drop grows with the square of the time from the peak. Divide the landing drop by the earlier drop to remove the same multiplier $k$ :
$\frac{k(9-v)^2}{k(1-v)^2}=\frac{180}{20}.$
Cancel $k$ :
$\frac{(9-v)^2}{(1-v)^2}=\frac{180}{20}.$
Cross-multiply:
$20(9-v)^2=180(1-v)^2.$
Step 5
Find when the rocket peaks
Type $20(9-v_1)^2\sim180(1-v_1)^2\{0\le v_1<9\}$ in Desmos. The $\sim$ asks Desmos to solve for $v_1$ ; the restriction keeps the peak between launch and landing. Under PARAMETERS, Desmos shows $v_1=3$ , so the rocket peaks $3$ seconds after launch.
Step 6
Find the multiplier
Now put $v=3$ into the $20$ -foot drop equation. Type $k(1-3)^2\sim20$ on the next Desmos line. Under PARAMETERS, Desmos shows $k=5$ .
Step 7
Evaluate the height at launch
Define $h(x)=180-k(x-v_1)^2$ in Desmos, where $x$ stands for time, then type $h(0)$ . Desmos prints $135$ . The input is $0$ at launch, so the rocket started $135$ feet above the ground. Choice B.

Question 61·200 Super-Hard SAT Math Questions·Advanced Math

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