A quadratic function models a projectile’s height, in meters, above the ground seconds after it is launched. The model estimates that: - The projectile’s maximum height occurs at . - . - . According to the model, what is the difference, in seconds, between the two times when the projectile’s height is meters?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 60 | Free SAT Prep | Aniko

00:00

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

0 / 200

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?

When a quadratic’s maximum (or minimum) time is given, start in vertex form so the axis of symmetry is built in. Then use the given function values to eliminate unnecessary constants by subtraction (here, subtracting cancels both $p$ and $a$ ), solve for the remaining parameter, and finally solve for the two symmetric times. If the question asks for the difference between the two times, compute the spacing directly rather than reporting either time.

Hints

Click me!

Use the location of the maximum

A quadratic with a maximum at $t=5$ can be written as $h(t)=p-k(t-5)^2$ for some constants $p$ and $k$ .

Eliminate $p$ and $a$

Write equations for $h(0)$ and $h(3)$ . Subtract them to eliminate both $p$ and $a$ so you can solve for $k$ .

Use symmetry after solving

When you solve $h(t)=a+8$ , you should get two solutions of the form $5\pm d$ . The difference between the times is $2d$ .

Desmos Guide

Use a simplified model with $a=0$

Because $a$ cancels when comparing heights like $a$ , $a+32$ , and $a+8$ , set $a=0$ to simplify. Define

$k=32/21$
$p=25k$

in Desmos.

Graph the height function and the target height

Enter

$y=p-k(x-5)^2$
$y=8$

where $x$ represents time $t$ .

Find the two intersection times

Click the two intersection points of the graphs and record their $x$ -coordinates, say $t_1$ and $t_2$ .

Compute the difference

In a new expression line, enter $t_2-t_1$ (using the recorded values) and simplify the result. That value is the requested difference.

Step-by-step Explanation

Write the quadratic in vertex form

Since the maximum height occurs at $t=5$ , write

h(t)=p-k(t-5)^2

where $k>0$ and $p$ is the maximum height.

Use the given heights to find $k$

From $h(0)=a$ :

p-k(0-5)^2=p-25k=a

From $h(3)=a+32$ :

p-k(3-5)^2=p-4k=a+32

Subtract the first equation from the second:

(p-4k)-(p-25k)=(a+32)-a

21k=32 \quad\Rightarrow\quad k=\frac{32}{21}

Set the height equal to $a+8$ and solve for the times

First find $p$ from $p-25k=a$ :

p=a+25k

Now solve $h(t)=a+8$ :

(a+25k)-k(t-5)^2=a+8

Cancel $a$ and divide by $k$ :

25-(t-5)^2=\frac{8}{k}

Substitute $k=\frac{32}{21}$ :

25-(t-5)^2=8\cdot\frac{21}{32}=\frac{21}{4}

(t-5)^2=25-\frac{21}{4}=\frac{79}{4}

which gives two times:

t=5\pm\frac{\sqrt{79}}{2}

Find the difference between the two times

The two times are $5-\frac{\sqrt{79}}{2}$ and $5+\frac{\sqrt{79}}{2}$ , so their difference is

\left(5+\frac{\sqrt{79}}{2}\right)-\left(5-\frac{\sqrt{79}}{2}\right)=\sqrt{79}

Therefore, the difference between the two times is $\sqrt{79}$ seconds.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation