Question 60·200 Super-Hard SAT Math Questions·Advanced Math

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A quadratic function $h$ models a projectile’s height, in meters, above the ground $t$ seconds after it is launched.

The model estimates that:

The projectile’s maximum height occurs at $t=5$ .
$h(0)=a$ .
$h(3)=a+32$ .

According to the model, what is the difference, in seconds, between the two times when the projectile’s height is $a+8$ meters?

A parabola is a mirror-image curve about its peak, so equal heights occur equally far before and after the maximum. Use a known height to find its mirrored time, then build a quadratic for the height above that level. Solve for each time’s distance from the peak. The gap is twice that distance, not the time of the peak.

Hints

Hint 1
The maximum at $t=5$ is the center of the parabola’s mirror image. Launch at $t=0$ is $5$ seconds before the maximum, so at what time is the projectile equally far after it?
Hint 2
Subtract the starting height $a$ . The remaining height is $0$ at launch and at its mirrored time, so its quadratic expression has a factor that becomes zero at each of those times. Use $h(3)$ to find the multiplier.
Hint 3
Call the distance from the peak to either requested time $d$ . Those times are $5-d$ and $5+d$ , so the time gap is $2d$ . What equation makes their height $8$ meters above $a$ ?

Step-by-step

Use symmetry and distance from the peak
Step 1
Find the matching time at height $a$
The peak at $t=5$ is the axis of symmetry, the mirror line of the parabola. Launch at $t=0$ is $5$ seconds before it, so $5$ seconds after it is $t=10$ . Mirror times have equal heights, so:
$h(10)=h(0)=a.$
Step 2
Write the height above $a$
The difference $h(t)-a$ is zero at $t=0$ and $t=10$ . A quadratic with those zeros has a factor for each time:
$h(t)-a=kt(10-t),$
where $k$ is a constant that sets the size of the height change.
Step 3
Find the multiplier
The given $h(3)=a+32$ puts the projectile $32$ meters above $a$ at $t=3$ . Substitute into the model:
$32=k(3)(10-3).$
Type $32\sim k(3)(10-3)$ in Desmos to find $k$ . Under PARAMETERS, it shows $k\approx1.52381$ . Simplify the product:
$32=21k.$
Divide by $21$ :
$k=\frac{32}{21}.$
Step 4
Measure time from the peak
Let $d>0$ be the distance in seconds from the peak to either time at the requested height. The times are $5-d$ and $5+d$ . For either time, substitute into the model’s product:
$t(10-t)=(5-d)(5+d).$
Multiply the factors:
$t(10-t)=25-d^2.$
Step 5
Set the height gain equal to $8$
A height of $a+8$ is $8$ meters above $a$ , so:
$\frac{32}{21}(25-d^2)=8.$
Type $\frac{32}{21}(25-d_1^2)\sim8\{d_1>0\}$ in Desmos. The $\sim$ asks it to fit $d_1$ ; the restriction keeps the positive distance. Under PARAMETERS, Desmos shows $d_1\approx4.4441$ seconds. That’s the distance on one side of the peak, not the gap.
Step 6
Find the exact distance
Keep the distance exact. Divide the equation by $\frac{32}{21}$ ; type $\frac{8}{\frac{32}{21}}$ below the regression. Desmos displays $5.25$ , or $\frac{21}{4}$ using its fraction button:
$25-d^2=\frac{21}{4}.$
Subtract $25=\frac{100}{4}$ :
$-d^2=-\frac{79}{4}.$
Multiply by $-1$ :
$d^2=\frac{79}{4}.$
Take the positive square root, since $d$ is a distance:
$d=\sqrt{\frac{79}{4}}.$
Step 7
Find the time gap
Subtract the earlier time from the later one:
$(5+d)-(5-d)=2d.$
Substitute the exact distance:
$2d=2\left(\sqrt{\frac{79}{4}}\right).$
Simplify the square root of $4$ :
$2d=2\left(\frac{\sqrt{79}}{2}\right).$
Cancel the $2$ :
$2d=\sqrt{79}.$
Type $2d_1$ on the next Desmos line; it prints about $8.8882$ seconds. The difference between the two times is exactly $\sqrt{79}$ seconds. Choice A.

Question 60·200 Super-Hard SAT Math Questions·Advanced Math

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