Projectile Motion with Parametric Equations

Projectile motion with parametric equations models a thrown or launched object with one equation for horizontal distance and one for height. With launch speed $v_0$ , angle $\theta$ and starting height h, in feet, $x = (v_0\cos\theta)t$ and $y = h + (v_0\sin\theta)t - 16t^2$ . In meters, replace 16 with 4.9.

Before this lesson: Parametric Equations, Magnitude and Direction of a Vector and Quadratic Formula

Why a projectile needs two equations

When you throw a ball, two things happen at once. It moves forward at a steady speed, and it rises and then falls because gravity pulls it down. Parametric equations let us handle these two motions separately, with time t as the parameter.

The launch velocity is a vector. Its horizontal part is $v_0\cos\theta$ and its vertical part is $v_0\sin\theta$ , just like the components of any vector. Gravity only changes the vertical part.

These equations ignore air resistance. That is a good model for a short throw and a rough one for a fast golf ball.

The path of a ball thrown at 80 ft/s at 30° from a height of 5 ft, graphed from x = 69.28t and y = 5 + 40t − 16t². It starts at (0, 5), reaches its highest point (86.6, 30) at t = 1.25 seconds, and lands at about (181, 0) at t ≈ 2.62 seconds. Dots mark the position every half second. — The ball from Examples 1–3. The dots are evenly spaced left to right, since the horizontal speed never changes.

Projectile motion equations

Units	Horizontal position	Height
Feet and seconds	$x = (v_0\cos\theta)t$	$y = h + (v_0\sin\theta)t - 16t^2$
Meters and seconds	$x = (v_0\cos\theta)t$	$y = h + (v_0\sin\theta)t - 4.9t^2$

Here $v_0$ is the launch speed, $\theta$ is the launch angle above the horizontal, and h is the launch height. The 16 and 4.9 are half of the acceleration due to gravity, 32 ft/s² or 9.8 m/s².

The height equation is a quadratic in t, so its graph over time is a parabola. Use these tools:

Question	How to answer it
When is the object highest?	At the vertex, $t = -\frac{b}{2a}$ for $y = at^2 + bt + c$
How high does it go?	Substitute that t into the height equation
When does it land?	Set y = 0 and solve, often with the quadratic formula
How far does it go?	Substitute the landing time into the x-equation
Does it clear an obstacle?	Solve x = (distance) for t, then find y at that t

Projectile motion examples

Example 1: Write the parametric equations

A ball is thrown at 80 feet per second at an angle of 30° from a height of 5 feet. Write parametric equations for its path.

x = (80\cos 30^\circ)t

Use $x = (v_0\cos\theta)t$ with $v_0 = 80$ and $\theta = 30^\circ$ .

x = 40\sqrt{3}\,t

Use $\cos 30^\circ = \frac{\sqrt{3}}{2}$ (see trig values of special angles).

x \approx 69.28t

Change to a decimal.

y = 5 + (80\sin 30^\circ)t - 16t^2

Use the height equation in feet with h = 5.

y = 5 + 40t - 16t^2

Use $\sin 30^\circ = \frac{1}{2}$ .

Answer: $x \approx 69.28t$ and $y = 5 + 40t - 16t^2$ .

Example 2: Maximum height

For the ball in Example 1, when does it reach its highest point, and how high is it?

y = -16t^2 + 40t + 5

Write the height in standard form: a = −16, b = 40, c = 5.

t = -\frac{40}{2(-16)}

The highest point is at the vertex, $t = -\frac{b}{2a}$ (see graphing quadratic functions).

t = 1.25

Simplify: −40 divided by −32.

y = 5 + 40(1.25) - 16(1.25)^2

Substitute t = 1.25 into the height equation.

y = 5 + 50 - 25

Multiply. Note $(1.25)^2 = 1.5625$ and $16(1.5625) = 25$ .

y = 30

Add and subtract.

x \approx 69.28(1.25) \approx 86.6

Find the horizontal distance at that time.

Answer: the ball is highest at t = 1.25 seconds, 30 feet up and about 86.6 feet out.

Example 3: Time in the air and distance

When does the ball from Example 1 hit the ground, and how far away does it land?

-16t^2 + 40t + 5 = 0

The ball lands when its height is 0.

t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(5)}}{2(-16)}

Use the quadratic formula with a = −16, b = 40, c = 5.

t = \frac{-40 \pm \sqrt{1600 + 320}}{-32}

Simplify under the root and in the denominator.

t = \frac{-40 \pm \sqrt{1920}}{-32}

Add.

t \approx \frac{-40 - 43.818}{-32}

Use the minus sign. The plus sign gives a negative time, which does not fit the problem.

t \approx 2.619

Divide.

x \approx 69.28(2.619) \approx 181

Substitute the landing time into the x-equation.

Answer: the ball lands after about 2.62 seconds, about 181 feet away.

Example 4: Does the ball clear the fence?

A batter hits a ball at 120 ft/s at an angle of 35°, from 3 feet above the ground. The outfield fence is 10 feet tall and 400 feet away. Does the ball clear it?

x = (120\cos 35^\circ)t \approx 98.298t

Write the horizontal equation.

y = 3 + (120\sin 35^\circ)t - 16t^2 \approx 3 + 68.829t - 16t^2

Write the height equation.

98.298t = 400

Find when the ball reaches the fence: set x = 400.

t \approx 4.0692

Divide both sides by 98.298.

y \approx 3 + 68.829(4.0692) - 16(4.0692)^2

Substitute that time into the height equation.

y \approx 3 + 280.08 - 264.93

Multiply.

y \approx 18

Add and subtract, and round to the nearest foot.

Answer: yes. When the ball reaches the fence it is about 18 feet high, about 8 feet above the 10-foot fence. It’s a home run.

Example 5: Eliminate the parameter to find the path

Write the path from Example 1 as one equation in x and y, using the exact form $x = 40\sqrt{3}\,t$ .

t = \frac{x}{40\sqrt{3}}

Solve the x-equation for t.

y = 5 + 40\left(\frac{x}{40\sqrt{3}}\right) - 16\left(\frac{x}{40\sqrt{3}}\right)^2

Substitute into the height equation.

y = 5 + \frac{x}{\sqrt{3}} - \frac{16x^2}{4800}

Simplify each term: $(40\sqrt{3})^2 = 1600 \cdot 3 = 4800$ .

y = 5 + \frac{x}{\sqrt{3}} - \frac{x^2}{300}

Reduce $\frac{16}{4800} = \frac{1}{300}$ .

Answer: $y = -\frac{x^2}{300} + \frac{x}{\sqrt{3}} + 5$ . The path is a downward parabola, which is why a thrown ball arcs the way it does. The parametric form also tells you when the ball is at each point; this form does not.

Common mistakes

Mistake	Example of the error	Fix
Using the full speed for both directions	$x = 80t$ for a throw at 30°	Split the speed: $x = (80\cos 30^\circ)t$ .
Using −32t² instead of −16t²	$y = 5 + 40t - 32t^2$	The term is half of 32: $-16t^2$ (or $-4.9t^2$ in meters).
Mixing feet and meters	Speed in m/s with −16t²	Use −4.9t² with meters and −16t² with feet.
Keeping the negative root for landing time	t ≈ −0.12 s	Time after launch must be positive, so keep t ≈ 2.62.
Forgetting the starting height	$y = 40t - 16t^2$ for a throw from 5 ft	Add h: $y = 5 + 40t - 16t^2$ .
Calculator in radian mode	$\cos 30 \approx 0.154$	Use degree mode: $\cos 30^\circ \approx 0.866$ .

Related methods

Quick formulas for a launch from the ground

When h = 0, the object lands at $t = \frac{2v_0\sin\theta}{g}$ , where g is 32 ft/s² or 9.8 m/s². You get this by factoring t out of $(v_0\sin\theta)t - \frac{1}{2}gt^2 = 0$ (see factoring out the GCF).

Height over time alone

If a problem only asks about height, you only need the y-equation. This is the same model used in quadratic equation word problems about objects thrown straight up.

Practice problems

A ball is thrown at 60 ft/s at 45° from a height of 6 feet. Write parametric equations for its path.
A ball follows $x = 50t$ and $y = 4 + 30t - 16t^2$ , in feet. Where is it at t = 2? What is happening then?
When does $y = 3 + 48t - 16t^2$ reach its maximum height, and what is that height?
A ball’s height is $y = 64t - 16t^2$ . When does it hit the ground?
An arrow is shot at 40 m/s at 20° from 1.5 meters high. Write parametric equations, in meters.
A rock is launched from the ground at 30 m/s at 60°. How long is it in the air?
How far does the rock in problem 6 travel horizontally?
A golf ball is hit from the ground at 150 ft/s at 12°. How high does it go?
Eliminate the parameter: $x = 50t$ and $y = 4 + 30t - 16t^2$ .
A ball is hit at 100 ft/s at 40° from 3 feet high. Does it clear a 12-foot wall 330 feet away?

Show answers

x = (60\cos 45^\circ)t

Write the horizontal equation.

x = 30\sqrt{2}\,t \approx 42.43t

Use $\cos 45^\circ = \frac{\sqrt{2}}{2}$ .

y = 6 + (60\sin 45^\circ)t - 16t^2

Write the height equation with h = 6.

y = 6 + 30\sqrt{2}\,t - 16t^2 \approx 6 + 42.43t - 16t^2

Use $\sin 45^\circ = \frac{\sqrt{2}}{2}$ .

Answer: $x \approx 42.43t$ , $y \approx 6 + 42.43t - 16t^2$

x = 50(2) = 100

Substitute t = 2 into the x-equation.

y = 4 + 30(2) - 16(2)^2

Substitute t = 2 into the y-equation.

y = 4 + 60 - 64 = 0

Multiply, then add and subtract.

Answer: (100, 0). The height is 0, so the ball hits the ground 100 feet away at t = 2.

t = -\frac{48}{2(-16)} = 1.5

The vertex is at $t = -\frac{b}{2a}$ .

y = 3 + 48(1.5) - 16(1.5)^2

Substitute t = 1.5.

y = 3 + 72 - 36 = 39

Multiply, then add and subtract.

Answer: at t = 1.5 seconds, 39 feet high.

64t - 16t^2 = 0

Set the height equal to 0.

16t(4 - t) = 0

Factor out 16t.

t = 0 \text{ or } t = 4

Set each factor equal to 0 (see solving quadratics by factoring). t = 0 is the launch.

Answer: it lands at t = 4 seconds.

x = (40\cos 20^\circ)t \approx 37.59t

Write the horizontal equation.

y = 1.5 + (40\sin 20^\circ)t - 4.9t^2

Write the height equation in meters with h = 1.5.

y \approx 1.5 + 13.68t - 4.9t^2

Find the sine in degree mode and multiply.

Answer: $x \approx 37.59t$ , $y \approx 1.5 + 13.68t - 4.9t^2$

y = (30\sin 60^\circ)t - 4.9t^2 \approx 25.981t - 4.9t^2

Write the height equation, with h = 0.

25.981t - 4.9t^2 = 0

Set the height equal to 0.

t(25.981 - 4.9t) = 0

Factor out t.

t \approx \frac{25.981}{4.9} \approx 5.30

Solve $25.981 - 4.9t = 0$ (t = 0 is the launch).

Answer: about 5.30 seconds

x = (30\cos 60^\circ)t = 15t

Write the horizontal equation, using $\cos 60^\circ = \frac{1}{2}$ .

x \approx 15(5.302) \approx 79.5

Substitute the landing time from problem 6.

Answer: about 79.5 meters

v_y = 150\sin 12^\circ \approx 31.19

Find the vertical part of the launch speed.

t \approx \frac{31.19}{32} \approx 0.975

The vertex of $y = 31.19t - 16t^2$ is at $t = -\frac{b}{2a} = \frac{31.19}{32}$ .

y \approx 31.19(0.975) - 16(0.975)^2

Substitute that time into the height equation.

y \approx 30.41 - 15.21 \approx 15.2

Multiply, then subtract.

Answer: about 15.2 feet

t = \frac{x}{50}

Solve the x-equation for t.

y = 4 + 30\left(\frac{x}{50}\right) - 16\left(\frac{x}{50}\right)^2

Substitute into the y-equation.

y = 4 + 0.6x - \frac{16x^2}{2500}

Simplify each term.

y = 4 + 0.6x - 0.0064x^2

Divide: 16 ÷ 2500 = 0.0064.

Answer: $y = -0.0064x^2 + 0.6x + 4$

10.

x \approx 76.604t

$x = (100\cos 40^\circ)t$

y \approx 3 + 64.279t - 16t^2

$y = 3 + (100\sin 40^\circ)t - 16t^2$

76.604t = 330

Set x = 330, the distance to the wall.

t \approx 4.308

Divide both sides by 76.604.

y \approx 3 + 64.279(4.308) - 16(4.308)^2

Find the height at that time.

y \approx 3 + 276.91 - 296.94 \approx -17.0

Multiply, then add and subtract.

Answer: no. The height would be negative, which means the ball has already landed before it reaches the wall.

Projectile motion FAQ

What are the parametric equations for projectile motion?

In feet, $x = (v_0\cos\theta)t$ and $y = h + (v_0\sin\theta)t - 16t^2$ . In meters, use $-4.9t^2$ instead of $-16t^2$ .

How do you find the maximum height of a projectile?

Find the vertex of the height equation, $t = -\frac{b}{2a}$ , then substitute that time into y.

How do you find how far a projectile travels?

Set y = 0 and solve for the positive time. Substitute that time into the x-equation.

Why is the path of a projectile a parabola?

x grows at a steady rate while y is a quadratic in t. Eliminating t gives y as a quadratic in x, and the graph of a quadratic is a parabola.

Summary

Split the launch speed into $v_0\cos\theta$ (horizontal) and $v_0\sin\theta$ (vertical).
Feet: $y = h + (v_0\sin\theta)t - 16t^2$ . Meters: $y = h + (v_0\sin\theta)t - 4.9t^2$ .
Highest point: the vertex, $t = -\frac{b}{2a}$ .
Landing: set y = 0 and keep the positive root; distance: put that t into x.
To test an obstacle, solve x = distance for t, then find y.

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