Projectile motion SAT Desmos: Quadratic Word Problems, Max Height, and Landing Time

By the Cheetah Prep team · Reviewed July 13, 2026

For projectile motion SAT Desmos questions, the fastest calculator method is to type the height function into Desmos, then use the graph to pull the values you need, usually the maximum height and when the object hits the ground. This works because these word problems usually give a quadratic, and Desmos can show its key points right away.

A typical setup looks like $h(t)=at^2+bt+c$ , where $t$ is time and $h$ is height. In Desmos, you can:

Find max height SAT: tap the highest point of the parabola and read the vertex coordinates. The $y$ value is the maximum height, and the $x$ value is the time it happens.
Find when it lands: find the $x$ intercept where $h(t)=0$ (ignore negative time).
Answer “when is it at a certain height”: graph $h(t)$ and the horizontal line $y=k$ , then read the intersection times.

Before you trust the result, check that your variable and units match the question (time versus horizontal distance is a common trap). If you want a quick check that your final value actually fits the equation in the prompt, use the same idea as in How to check sat answers desmos by graphing both sides: graph what the problem says, then confirm the intersection or highlighted point matches your answer.

When to use this Desmos method

Use this Desmos method when the word problem is really about a quadratic’s key features, especially the vertex (maximum height) and the zeros (when it hits the ground).

It works best when the prompt gives you, or lets you write, a height function like $h(t)=at^2+bt+c$ , then asks for a value you can read from the graph.

Look for these SAT patterns:

Projectile or arc language: thrown, launched, kicked, fired, jumps, falls, path, height after $t$ seconds.
Maximum or minimum questions: highest point, greatest height, max height sat, peak, greatest value of $h(t)$ , when the height is greatest.
Landing or hitting the ground: when height is $0$ , when it lands, when it returns to the ground, time to hit the ground (choose the nonnegative intercept).
Same height twice: when is it at $k$ feet or meters, at what times is the height $k$ (you will usually see two intersection times).
Compare two models quickly: two different launch equations, or two objects, and the question is which goes higher or lands sooner, or has a larger maximum.

This is a good pick when the algebra is annoying, like messy coefficients, or when you would otherwise need completing the square or the quadratic formula.

Do not use it if the model is not a quadratic (piecewise rules, bounce stories, air resistance) or if the variable is not time. First confirm what the $x$ axis represents, then graph it. To check your final choice fast, use SAT Desmos guides to practice quick graph based checks.

Step by step in Desmos

Define the variables the question is using
Read the prompt and decide what the input means. Most projectile motion SAT Desmos questions use time as the input, so use $t$ for time and $h(t)$ for height. If the prompt uses $x$ instead, match that to avoid mixing units.
Type the height model as a quadratic
Enter the quadratic height function exactly as given, or build it from the information in the prompt. Use parentheses carefully so the coefficient applies to the entire squared term.
h(t)=at^2+bt+c
Set a realistic viewing window
If the graph looks flat or you cannot see the peak, adjust the window. You want to see from the launch time through the landing time, and from height $0$ up past the highest point. This prevents you from tapping the wrong point.
Grab the maximum height from the vertex
Tap the highest point of the parabola. Desmos shows the vertex coordinates. The $y$ value is the maximum height, and the $x$ value is the time when the max height happens. This is the fastest way to answer max height sat questions.
Find when it hits the ground using the x intercept
Find where the graph crosses the horizontal axis, where $h(t)=0$ . Tap that intercept and read the $x$ value as the landing time. If there are two intercepts, ignore the negative time and keep the nonnegative one.
Answer a specific height question with a horizontal line
If the question says when the object is at height $k$ , add the line $y=k$ . The intersection points with $h(t)$ give the times. You will often get two times, one on the way up and one on the way down.
y=k
Use a quick restriction if negative time is distracting
If the left side of the parabola is cluttering the view, restrict the graph to nonnegative time so you only see the physically meaningful part. Then your taps automatically land on valid times.
h(t){t>=0}

Exact expressions to enter

$h(t)=at^2+bt+c$ Type this into Desmos
Type the height function from the prompt. Use t for time if the prompt uses time.
$h(t)=(-g/2)t^2+v_0t+h_0$ Type this into Desmos
Generic projectile model. Replace g, v_0, and h_0 with the values the problem gives. If the prompt uses different letters, match them.
$y_1=h(t)$ Type this into Desmos
Optional: name the function so later expressions are easier to read.
$y=0$ Type this into Desmos
Ground line. Use this to find when it lands by reading the nonnegative intersection with h(t).
$y=k$ Type this into Desmos
Use when the question asks for the time when the object is at height k. Read the intersection x values.
$v(t)=h'(t)$ Type this into Desmos
Optional: velocity as the derivative. The peak happens when v(t)=0.
$v(t)=0$ Type this into Desmos
Optional: solve for the time of maximum height by finding where the derivative hits 0.
$t_{peak}=-b/(2a)$ Type this into Desmos
Optional: if you entered h(t)=at^2+bt+c, this gives the vertex time directly.
$h(t_{peak})$ Type this into Desmos
Optional: evaluate the maximum height once you have t_{peak}.
$h(t)=0$ Type this into Desmos
Optional: if you prefer a solve style, graph this and read the solutions for landing times.
$t\ge0$ Type this into Desmos
Time restriction reminder. Ignore negative time solutions.

Worked SAT style example

Example

A ball is launched upward from a platform. Its height in meters after $t$ seconds is modeled by $h(t)=-4.9t^2+14t+2$ . Using Desmos, find the maximum height of the ball and the time when it hits the ground.

Type $h(t)=-4.9t^2+14t+2$ into Desmos.
Find the maximum height SAT value: tap the highest point of the parabola (the vertex). Read the coordinates. The $y$ value is the maximum height, and the $x$ value is the time when the height is greatest.
Find when it hits the ground: look for where the graph crosses the $t$ axis. If you do not see the intercept clearly, type $h(t)=0$ on a new line and use the intersection point. Choose the intercept with $t\ge 0$ .
Sanity check: the equation opens downward because the $t^2$ coefficient is negative, so there should be a single peak and two zeros, one negative time and one nonnegative time. The landing time must be the nonnegative one.

Answer: Maximum height is

12

meters at

t=\frac{10}{7}

seconds. The ball hits the ground at

t=\frac{10+\sqrt{170}}{7}

seconds.

Common mistakes

Most wrong answers on projectile motion SAT Desmos questions happen because you read the graph right, then build the wrong equation or grab the wrong point.

Using the wrong input variable: Many prompts use $t$ for time, but some use $x$ for horizontal distance. If you type $h(x)$ when the equation is really $h(t)$ , your vertex and intercepts will not match what the question asks.
Forgetting that only nonnegative time makes sense: Desmos will show two $x$ intercepts for many parabolas. The negative one is not a landing time, even if it looks clean.
Reading the wrong coordinate at the peak: The maximum height is the vertex $y$ value, not the $x$ value. The $x$ value is the time when max height happens.
Mixing units inside the model: If the prompt gives seconds for time but your equation uses minutes, or gives feet for height but you treat them like meters, the graph will still look like a parabola but the numbers will be off. For a fast fix, use the canceling factor idea from Desmos Unit Conversion SAT: Fast Canceling Factor Method.
Entering the quadratic with missing parentheses: Type $h(t)=at^2+bt+c$ carefully. For example, $-(t-3)^2+10$ is very different from $-t-3^2+10$ .
Solving the wrong question: Some prompts ask for “time to reach max height,” others ask for “maximum height.” In Desmos, those are different parts of the vertex.
Ignoring “same height twice”: If you graph $y=k$ , you will often get two intersection times. Do not automatically pick the smaller one. Read the prompt for which event it describes.

When this method does not work

This method fails when the prompt is not actually a quadratic in one variable, or when the question depends on restrictions that the graph will not enforce.

Watch out for these cases:

The model is not quadratic: If height changes with air resistance, bouncing, or a piecewise rule, $h(t)$ might not be $at^2+bt+c$ . Desmos can still graph it, but the vertex and intercepts you read off will not match the situation.
Your variable is wrong: Some problems give height as a function of horizontal distance, not time. If you graph $h(t)$ but the prompt is really $h(x)$ , the maximum height and landing point you read off answer the wrong question.
The question needs an exact value: Desmos often shows a decimal. If the SAT wants a simplified radical or a clean fraction, you will need algebra to finish, even if the graph got you close.
Domain matters: You must restrict to physically meaningful inputs, usually $t \ge 0$ . Desmos will still show negative time intercepts, or a second intersection after the object has already hit the ground, if the model only applies up to landing.
Units are mixed: If the prompt mixes feet, meters, or seconds and minutes, graphing the equation as written can give the wrong result. Convert first, for example in Desmos Unit Conversion SAT: Fast Canceling Factor Method.

Practice questions

1.A ball’s height in meters after $t$ seconds is modeled by $h(t)=-5t^2+20t+1$ . Using Desmos, what is the maximum height of the ball?

2.A toy rocket’s height in feet after $t$ seconds is $h(t)=-16t^2+48t+5$ . According to the model, at what time does the rocket hit the ground?

3.A stone is thrown upward from a platform. Its height in meters after $t$ seconds is $h(t)=-4t^2+12t+8$ . For what value of $t$ is the stone at a height of $12$ meters?

4.A basketball’s height in meters after $t$ seconds is $h(t)=-2t^2+8t+2$ . What is the time when the ball reaches its maximum height?

5.Two objects are launched. Their heights in meters after $t$ seconds are $h_1(t)=-t^2+6t$ and $h_2(t)=-2t^2+10t$ . Using Desmos, which object has the greater maximum height?

FAQ

What is the fastest Desmos approach for projectile motion SAT questions?

Type the height function $h(t)$ into Desmos. Use the graph to read the vertex for maximum height and the $x$ intercept for when it hits the ground. This is fast because both points are standard features of a quadratic graph, so you can skip completing the square and the quadratic formula unless the question demands it.

How do I find the maximum height SAT value in Desmos?

Graph $h(t)$ . Tap the vertex and read its coordinates. The $y$ coordinate is the maximum height. The $x$ coordinate is the time when that maximum happens.

How do I find when the object hits the ground in Desmos?

Find where the graph crosses the $t$ axis. That point is where $h(t)=0$ . If you see two intercepts, pick the nonnegative time. If the model starts on the ground at $t=0$ , one intercept is often $t=0$ . The other intercept is the landing time.

The question asks when it is at a certain height. What should I graph?

Graph the quadratic $h(t)$ . Then graph the horizontal line $y=k$ , where $k$ is the height in the question. The intersection points are the times. You usually get two times because the object goes up, then comes back down.

What if the equation uses $x$ instead of $t$?

Use the variable the problem uses. If the model is $h(x)$ , Desmos will still show you a parabola, but the horizontal axis means whatever the problem defines, time or horizontal distance. Most mistakes in projectile word problems come from mixing up units or what the variable represents.

Do I need to rewrite the quadratic in vertex form to get max height?

No. Desmos shows the vertex straight from standard form $h(t)=at^2+bt+c$ . Vertex form can help on algebra only questions. On calculator active questions, reading the vertex off the graph is usually faster and less error prone.

My graph shows a negative maximum height or the vertex is not between the intercepts. What went wrong?

Check three things: you entered the signs correctly, you used the right variable, and you did not mix units. A projectile height model with a maximum opens downward, so the coefficient on $t^2$ should be negative. If it opens upward, you probably flipped a sign, or you are not modeling height.

How do I handle answer choices that are times, but Desmos gives a decimal?

Use the decimal you get from the intercept or intersection, then pick the closest answer choice. If the choices are exact expressions, use Desmos to check them: plug in each one as $h(\text{choice})$ and see which one makes the statement true, like $h(t)=0$ or $h(t)=k$ .

What if the question gives a verbal description, not the full equation?

Write $h(t)$ first. Then graph it. Many SAT prompts give you enough to write a quadratic, for example an initial height and how the height changes with time. After you have a quadratic model, use Desmos to read off the maximum height and the landing time the same way.

Are projectile motion questions always about quadratics?

On the SAT, projectile style word problems that fit this Desmos method describe a parabolic arc, so they use a quadratic relationship. If the prompt points to a different model, use that model. Do not force a quadratic.

About this page: written and reviewed by the Cheetah Prep team. Last reviewed July 13, 2026.