Projectile Motion, Motion in Two Dimensions | Physics

Learning objectives

Teacher Support

The learning objectives in this section will help your students master the following standards:

(4) Science concepts. The student knows and applies the laws governing motion in two dimensions for a variety of situations. The student is expected to:
- (C) analyze and describe accelerated motion in two dimensions using equations.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:

(4) Science concepts. The student knows and applies the laws governing motion in a variety of situations. The student is expected to:
- (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

Section Key Terms

air resistance	maximum height (of a projectile)	projectile
projectile motion	range	trajectory

Properties of Projectile Motion

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile, and its path is called its trajectory. Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.

The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are independent, meaning that they don’t influence one another. Figure_05_03_cannonball compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.

Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.

View original figure (external; image not reproduced in this text edition)

Original figure description: A cannonball is sitting on a flat surface. Dotted lines show the projectile motion of a cannon ball. Dotted line A shows the ball dropping vertically. Dotted line B shows a curved path. Dotted line C shows the ball moving horizontally.

The diagram shows the projectile motion of a cannonball shot at a horizontal angle versus one dropped with no horizontal velocity. Note that both cannonballs have the same vertical position over time.

We’ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y, as illustrated in Figure_05_03_soccer.

View original figure (external; image not reproduced in this text edition)

Original figure description: A boy is kicking a soccer ball. A curved line shows the ball's motion. A right triangle is drawn with vectors using an x-y axis. The hypotenuse is labeled d, the vertical leg is labeled y and the horizontal leg is labeled x.

A boy kicks a ball at angle θ, and it is displaced a distance of s along its trajectory.

As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple a_y = –g = –9.80 m/s². Note that this definition defines the upwards direction as positive. Because gravity is vertical, a_x = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table_05_01.

Summary of Kinematic Equations (constant a)
$x = x_{0} + v_{a v g} t$ (when $a = constant$ )
$v_{a v g} = \frac{v_{0} + v}{2}$ (when $a = 0$ )
$v = v_{0} + a t$
$x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$
$v^{2} = v_{0}^{2} + 2 a (x - x_{0})$

Where x is position, x₀ is initial position, v is velocity, v_avg is average velocity, t is time and a is acceleration.

Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:

Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so $A_{x} = A cos θ$ and $A_{y} = A sin θ$ are used. The magnitudes of the displacement $s$ along x- and y-axes are called $x$ and $y .$ The magnitudes of the components of the velocity $v$ are $v_{x} = v \cos θ$ and $v_{y} = v \sin θ$ , where $v$ is the magnitude of the velocity and $θ$ is its direction. Initial values are denoted with a subscript 0.
Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms $\begin{matrix} Horizontal Motion (a_{x} = 0) \\ x = x_{0} + v_{x} t \\ v_{x} = v_{0 x} = v_{x} = velocity is a constant. \end{matrix}$ Vertical motion (assuming positive is up $a_{y} = - g = - 9.80 {m/s}^{2}$ ) $\begin{matrix} y & = & y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t \\ v_{y} & = & v_{0 y} - g t \\ y & = & y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} \\ v_{y}^{2} & = & v_{0 y}^{2} - 2 g (y - y_{0}) \end{matrix}$
Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time $t$ . The problem solving procedures here are the same as for one-dimensional kinematics.
Recombine the two motions to find the total displacement $s$ and velocity $v$ . We can use the analytical method of vector addition, which uses $A = \sqrt{A_{x}^{2} + A_{y}^{2}}$ and $θ = \tan^{- 1} (A_{y} / A_{x})$ to find the magnitude and direction of the total displacement and velocity. $\begin{matrix} Displacement \\ d = \sqrt{x^{2} + y^{2}} \\ θ = \tan^{- 1} (y / x) \\ Velocity \\ v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ θ_{v} = \tan^{- 1} (v_{y} / v_{x}) \end{matrix}$ $θ$ is the direction of the displacement $d$ , and $θ_{v}$ is the direction of the velocity $v$ . (See Figure_05_03_projectile

View original figure (external; image not reproduced in this text edition)

Original figure description: Four related diagrams of projectile motion are show, a, b, c, and d. Diagram a shows two-dimensional curved motion, vx and vy, along an x-y axis and includes the formulas V y equals negative V oy and angle equals negative angle o. Diagram b shows horizontal motion, vx. Diagram c shows upward and downward vertical velocity, vy. And diagram d shows the combination of vx, vy, and v, forming a right triangle.

(a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because $a_{x} = 0$ and $v_{x}$ is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x- and y-motions are recombined to give the total velocity at any given point on the trajectory.

A Fireworks Projectile Explodes High and Away

During a fireworks display like the one illustrated in Figure_05_03_firework, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

View original figure (external; image not reproduced in this text edition)

Original figure description: The diagram shows the trajectory of a fireworks shell. An x-axis is labeled x, a y-axis is labeled h, and a vector, labeled V o, points upward from the axis origin. An angle is formed by the vector and the x-axis with a measure of seventy-five degrees.

The diagram shows the trajectory of a fireworks shell.

Solution for (a)

By height we mean the altitude or vertical position $y$ above the starting point. The highest point in any trajectory, the maximum height, is reached when $v_{y} = 0$ ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v_y when the firework reaches its maximum height, we use the following equation to find $y$

v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0}) .

Because $y_{0}$ and $v_{y}$ are both zero, the equation simplifies to

0 = v_{0 y}^{2} - 2 g y .

Solving for $y$ gives

y = \frac{v_{0 y}^{2}}{2 g} .

Now we must find $v_{0 y}$ , the component of the initial velocity in the y-direction. It is given by $v_{0 y} = v_{0} \sin θ$ , where $v_{0}$ is the initial velocity of 70.0 m/s, and $θ = 75^{\circ}$ is the initial angle. Thus,

v_{0 y} = v_{0} \sin θ_{0} = (70.0 m/s) (\sin 75^{\circ}) = 67.6 m/s

and $y$ is

y = \frac{{(67.6 m/s)}^{2}}{2 (9.80 {m/s}^{2}),}

so that

y = 233 m .

Discussion for (a)

Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

Solution for (b)

There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use $y = y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t$ . Because $y_{0}$ is zero, this equation reduces to

y = \frac{1}{2} (v_{0 y} + v_{y}) t .

Note that the final vertical velocity, $v_{y}$ , at the highest point is zero. Therefore,

\begin{matrix} t & = & \frac{2 y}{(v_{0 y} + v_{y})} = \frac{2 (233 m)}{(67.6 m/s)} \\ = & 6.90 s. \end{matrix}

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using $y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$ , and solving the quadratic equation for $t$ .

Solution for (c)

Because air resistance is negligible, $a_{x} = 0$ and the horizontal velocity is constant. The horizontal displacement is horizontal velocity multiplied by time as given by $x = x_{0} + v_{x} t$ , where $x_{0}$ is equal to zero

x = v_{x} t,

where $v_{x}$ is the x-component of the velocity, which is given by $v_{x} = v_{0} \cos θ_{0} .$ Now,

v_{x} = v_{0} \cos θ_{0} = (70.0 m/s) (\cos 75^{\circ}) = 18.1 m/s .

The time $t$ for both motions is the same, and so $x$ is

x = (18.1 m/s) (6.90 s) = 125 m.

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the forces of the explosion.

The expression we found for $y$ while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height $y = h$ ; then,

h = \frac{v_{0 y}^{2}}{2 g} .

This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity.

Calculating Projectile Motion: Hot Rock Projectile

Suppose a large rock is ejected from a volcano, as illustrated in Figure_05_03_volcanic, with a speed of $25.0 m / s$ and at an angle $35 °$ above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

View original figure (external; image not reproduced in this text edition)

Original figure description: A volcano is shown with labels illustrating the projectile motion of rock shooting out of the volcano. Vector v o equals twenty-five meters per second and points from the volcano opening upward. Vector v curves downward to the ground and forms an angle of thirty-five degrees with a horizontal plane at the height of the volcano opening. The distance between the horizontal plane and the ground is labeled as negative twenty meters.

The diagram shows the projectile motion of a large rock from a volcano.

Solution

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} .

If we take the initial position $y_{0}$ to be zero, then the final position is $y = - 20.0 m .$ Now the initial vertical velocity is the vertical component of the initial velocity, found from

v_{0}_{y} = v_{0} sin θ_{0} = (25.0 m/s) (sin 35^{\circ}) = 14.3 m/s .

Substituting known values yields

- 20.0 m = (14.3 m/s) t - (4.90 {m/s}^{2}) t^{2} .

Rearranging terms gives a quadratic equation in $t$

(4.90 {m/s}^{2}) t^{2} - (14 .3 m/s) t - (20.0 m) = 0 .

This expression is a quadratic equation of the form $a t^{2} + b t + c = 0$ , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

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

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

t = 3.96 s.

Discussion

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of $14.3 m / s$ and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Practice Problems

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The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure_05_03_trajectory. Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.

View original figure (external; image not reproduced in this text edition)

Original figure description: Two trajectory diagrams are shown. Diagram a shows three trajectories that curve upward then downward. They are labeled as follows: thirty meters per second and ninety-eight point one meters; forty meters per second and one hundred sixty-three meters; fifty meters per second and two hundred fifty-five meters. Diagram b shows three trajectories that curve upward then downward. They are labeled as follows: fifteen degrees and one hundred twenty-eight meters; forty-five degrees and two hundred fifty-five meters; seventy-five degrees and one hundred twenty-eight meters. The diagram also includes the formula v o equals fifty meters per second.

Trajectories of projectiles on level ground. (a) The greater the initial speed

v_{0}

, the greater the range for a given initial angle. (b) The effect of initial angle

θ_{0}

on the range of a projectile with a given initial speed. Note that any combination of trajectories that add to 90 degrees will have the same range in the absence of air resistance, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed $v_{0}$ , the greater the range, as shown in the figure above. The initial angle $θ_{0}$ also has a dramatic effect on the range. When air resistance is negligible, the range $R$ of a projectile on level ground is

R = \frac{v_{0}^{2} \sin 2 θ_{0}}{g},

where $v_{0}$ is the initial speed and $θ_{0}$ is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

Section Summary

Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
Projectile motion in the horizontal and vertical directions are independent of one another.
The maximum height of an projectile is the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile.
The range is the maximum horizontal distance traveled by a projectile.
To solve projectile problems: choose a coordinate system; analyze the motion in the vertical and horizontal direction separately; then, recombine the horizontal and vertical components using vector addition equations.

Key Equations

angle of displacement	$θ = \tan^{- 1} (y / x)$
velocity	$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
angle of velocity	$θ_{v} = \tan^{- 1} (v_{y} / v_{x})$
maximum height	$h = \frac{v_{0 y}^{2}}{2 g}$
range	$R = \frac{v_{0}^{2} \sin 2 θ_{0}}{g}$

Check Your Understanding

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What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance?

The nuclear force
The gravitational force
The electromagnetic force
The contact force

Concept Items

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Problems

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A person wants to fire a water balloon cannon such that it hits a target 100 m away. If the cannon can only be launched at 45° above the horizontal, what should be the initial speed at which it is launched?

31.3 m/s
37.2 m/s
980.0 m/s
1,385.9 m/s

Critical Thinking

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An object is launched into the air. If the y-component of its acceleration is 9.8 m/s², which direction is defined as positive?

Vertically upward in the coordinate system
Vertically downward in the coordinate system
Horizontally to the right side of the coordinate system
Horizontally to the left side of the coordinate system

Test Prep Multiple Choice

After a projectile is launched in the air, in which direction does it experience constant, non-zero acceleration, ignoring air resistance?

The x direction
The y direction
Both the x and y directions
Neither direction

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A ball is thrown in the air at an angle of 40°. If the maximum height it reaches is 10 m, what must be its initial speed?

17.46 m/s
21.78 m/s
304.92 m/s
474.37 m/s

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Test Prep Short Answer

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In the equation for the maximum height of a projectile, what does $v_{0 y}$ stand for? $h = \frac{v_{0 y}^{2}}{2 g}$

Initial velocity in the x direction
Initial velocity in the y direction
Final velocity in the x direction
Final velocity in the y direction

True or False—Range is defined as the maximum vertical distance travelled by a projectile.

True
False

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Test Prep Extended Response

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Glossary

air resistance: a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero
maximum height (of a projectile): the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile
projectile: an object that travels through the air and experiences only acceleration due to gravity
projectile motion: the motion of an object that is subject only to the acceleration of gravity
range: the maximum horizontal distance that a projectile travels
trajectory: the path of a projectile through the air

Attribution. Physics, OpenStax, Rice University, and the contributing authors credited in the book. Licensed under Creative Commons Attribution 4.0 International. Source revision 56e71824e7 of openstax/osbooks-physics (2026-10-13). CC BY 4.0. Original source ↗. Text edition of a pinned CC BY 4.0 source revision of the OpenStax book, converted from CNXML by the EvidenceHelix importer. Original prose, learning objectives, tables, exercises, glossary and captions are preserved; images are not copied (figures link to the original source and keep their credits). No new content was added. Later revisions of this book can carry different licence terms and are not used.