The height , in meters, of a projectile seconds after it is launched is modeled by the equation for . According to the model, how many seconds after the projectile is launched does the projectile reach a height of meters?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 203 | Free SAT Prep | Aniko

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Question 203·Medium·Nonlinear Functions

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The height $h$ , in meters, of a projectile $t$ seconds after it is launched is modeled by the equation

h(t) = -5t^{2} + 20t + 1.5

for $0 \le t \le 5$ .

According to the model, how many seconds after the projectile is launched does the projectile reach a height of $21.5$ meters?

For questions asking when a projectile reaches a certain height, set the given height function equal to the target height and rearrange to get a quadratic equation equal to zero. Simplify the coefficients (divide out common factors) and look first for easy factoring, especially perfect square trinomials; if factoring is hard, use the quadratic formula. As a quick check or backup, you can also plug each answer choice into the original height equation to see which time gives the required height, but algebra is usually faster and more reliable.

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Use the height equation

You are given a formula for $h(t)$ and a specific height value, $21.5$ . How can you combine these to form an equation you can solve for $t$ ?

Form a quadratic equation

After you set $-5t^2 + 20t + 1.5$ equal to $21.5$ , move all terms to one side so the equation equals zero. What quadratic equation in $t$ do you get?

Solve the quadratic efficiently

Once you have the quadratic, see if it can be factored easily. If it looks like a perfect square trinomial, write it as $(t - a)^2$ and solve for $t$ .

Desmos Guide

Enter the height function

In Desmos, type the function as y = -5x^2 + 20x + 1.5. This graphs the projectile’s height (y) as a function of time (x).

Enter the target height

On a new line, type y = 21.5. This draws a horizontal line at the height you’re interested in.

Find the intersection

Look for the point where the parabola and the horizontal line intersect between $x = 0$ and $x = 5$ . Click that intersection point and read off the x-coordinate; that x-value is the time (in seconds) when the projectile is at $21.5$ meters.

Step-by-step Explanation

Set up the equation for the given height

We are told the height is modeled by

h(t) = -5t^2 + 20t + 1.5

and we want to know when the height is $21.5$ meters.

So set $h(t)$ equal to $21.5$ :

-5t^2 + 20t + 1.5 = 21.5.

Move all terms to one side

Subtract $21.5$ from both sides to get a quadratic equation equal to zero:

-5t^2 + 20t + 1.5 - 21.5 = 0

which simplifies to

-5t^2 + 20t - 20 = 0.

To make the leading coefficient positive, multiply both sides by $-1$ :

5t^2 - 20t + 20 = 0.

Now divide everything by $5$ to simplify:

t^2 - 4t + 4 = 0.

Factor the quadratic

Factor $t^2 - 4t + 4$ .

We look for two numbers that multiply to $4$ and add to $-4$ . Those numbers are $-2$ and $-2$ , so we can factor as

t^2 - 4t + 4 = (t - 2)(t - 2) = (t - 2)^2.

So the equation becomes

(t - 2)^2 = 0.

Solve and match to the answer choice

Solve $(t - 2)^2 = 0$ by taking the square root of both sides:

t - 2 = 0,

t = 2.

This value is in the given interval $0 \le t \le 5$ , so according to the model, the projectile reaches a height of $21.5$ meters 2 seconds after launch. That corresponds to answer choice B) 2.

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Step-by-step Explanation

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Step-by-step Explanation