The height, in feet, of a ball thrown upward from a platform is modeled by the function where is the time in seconds after the ball is thrown (ignore air resistance). At what time or times will the ball be 42 feet above the ground?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 200 | Free SAT Prep | Aniko

00:00

Question 200·Medium·Nonlinear Functions

0 / 234

The height, in feet, of a ball thrown upward from a platform is modeled by the function

h(t) = -16t^{2} + 64t + 10,

where $t$ is the time in seconds after the ball is thrown (ignore air resistance).

At what time or times will the ball be 42 feet above the ground?

For height or position problems modeled by a quadratic function, first set the function equal to the given height or position to create an equation. Rearrange it into standard form $at^2 + bt + c = 0$ , then simplify by dividing out common factors to make the numbers smaller. If the quadratic does not factor nicely, use the quadratic formula and simplify the radical carefully so your final expression matches one of the answer choices exactly. Finally, check that your solutions make sense in context (for time problems, discard negative values).

Hints

Click me!

Use the given height

You are told the ball is 42 feet above the ground. How can you use the height function $h(t)$ to represent this situation as an equation?

Get a quadratic equation in standard form

After setting $h(t)$ equal to 42, move all terms to one side so the equation is equal to 0. Can you divide by a common factor to simplify the coefficients?

Solve the quadratic

Once you have an equation of the form $t^2 - 4t + 2 = 0$ , what method can you use to find the values of $t$ ? If it does not factor nicely, consider using the quadratic formula.

Simplify carefully

After using the quadratic formula, you will have a square root. Simplify the radical and the fraction so your final expression can be matched exactly to one of the answer choices.

Desmos Guide

Graph the height function

In Desmos, enter the function as y = -16x^2 + 64x + 10. Here, $x$ represents time $t$ in seconds and $y$ represents height in feet.

Graph the target height

On a new line, enter y = 42 to represent a horizontal line at 42 feet above the ground.

Find the intersection points

Use the intersection tool (or tap/click where the parabola and horizontal line meet) to find the $x$ -coordinates of the intersection points. Those $x$ -values are the times when the ball is 42 feet high; match them to the corresponding exact form in the answer choices.

Step-by-step Explanation

Set the height equal to 42 feet

We are told the height is 42 feet, so set $h(t)$ equal to 42:

-16t^2 + 64t + 10 = 42.

This equation tells us when the ball is 42 feet above the ground.

Rearrange into standard quadratic form

Move all terms to one side and simplify:

\begin{aligned} -16t^2 + 64t + 10 &= 42 \\ -16t^2 + 64t - 32 &= 0. \end{aligned}

Now divide every term by $-16$ to make the equation simpler:

\begin{aligned} t^2 - 4t + 2 &= 0. \end{aligned}

This is a standard quadratic equation in $t$ .

Apply the quadratic formula

For a quadratic equation $at^2 + bt + c = 0$ , the quadratic formula is

t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Here, $a = 1$ , $b = -4$ , and $c = 2$ . Substitute these into the formula:

\begin{aligned} t &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)} \\ &= \frac{4 \pm \sqrt{16 - 8}}{2} \\ &= \frac{4 \pm \sqrt{8}}{2}. \end{aligned}

This expression gives the two times when the ball is 42 feet high, but we can simplify it further.

Simplify and match the answer choice

Simplify $\sqrt{8}$ :

\sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}.

Substitute back:

\begin{aligned} t &= \frac{4 \pm 2\sqrt{2}}{2} \\ &= 2 \pm \sqrt{2}. \end{aligned}

So, the ball is 42 feet above the ground at $t = 2 - \sqrt{2}$ seconds and $t = 2 + \sqrt{2}$ seconds, which corresponds to answer choice D.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation