A projectile is any object that once launched or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

Projectile motion, also known as parabolic motion, is an example of composition of motion in two dimensions: an u.r.m. on the horizontal axis and a u.a.r.m. on the vertical axis. In this section, we will study:

Shall we begin?

## Concept and representation

Projectile motion, also known as parabolic motion, consists in launching a body with a velocity that form an angle α with the horizontal. In the following figure, you can see a representation of the situation.

Parabolic Motion

This motion is characteristic of projectiles, moving objects being affected only by gravity. On the x-axis, the body moves at constant velocity v0x (u.r.m.) and in the y-axis with constant acceleration due to gravity (u.a.r.m.).
It is characterized by the fact that at the highest point of the trajectory, the velocity of the body is always v0x (there is no vy).

Projectile motion or parabolic motion is the result of the composition of a uniform rectilinear motion (horizontal urm) and a uniformly accelerated rectilinear motion of launching upward or downward (vertical uarm).

## Equations

The equations for projectile motion are:

• The equations for the u.r.m. on the x-axis

$x={x}_{0}+{v}_{x}·t$

• The equations for the u.a.r.m. on the y-axis

${v}_{y}={v}_{0y}+{a}_{y}·t$

$y={y}_{0}+{v}_{0y}·t+\frac{1}{2}·{a}_{y}·{t}^{2}$

Since, as we said above, the velocity forms an angle α with the horizontal, the x and y components are determined using the most common trigonometric relationships:

Decomposition of the velocity vector

Any vector, including the velocity, can be broken down is in 2 vectors, vx and vy, that have the same directions as the Cartesian axes. The magnitude of both vectors can be calculated from the angle that the vector forms with the horizontal through the expressions shown in the figure.

Finally, taking into consideration the above said, that y0 = H , x0 = 0, and that ay = -g , , we can rewrite the formulas as are shown in the following list. These are the final expressions for calculating kinematics magnitudes in projectile motion or parabolic motion:

• Position (m)
• Horizontal axis

$x={v}_{x}\cdot t={v}_{0}·\mathrm{cos}\left(\alpha \right)·t$

• Vertical axis

$y=\mathrm{H}+{\mathrm{v}}_{0\mathrm{y}}·\mathrm{t}-\frac{1}{2}·g·{t}^{2}=\mathrm{H}+{\mathrm{v}}_{0}·\mathrm{sin}\left(\mathrm{\alpha }\right)·\mathrm{t}-\frac{1}{2}·g·{t}^{2}$

• Velocity (m/s)
• Horizontal axis

${v}_{x}={v}_{0x}={v}_{0}·\mathrm{cos}\left(\alpha \right)$

• Vertical axis

${v}_{y}={v}_{0y}-g\cdot t={v}_{0}·\mathrm{sin}\left(\alpha \right)-g\cdot t$

• Acceleration (m/s2)
• Horizontal axis

${a}_{x}=0$

• Vertical axis

${a}_{y}=-g$

Experiment and Learn

Data
g = 9.8 m/s2 |   |

Projectile motion

The blue ball in the figure represents a body suspended above the ground. You can drag it to the initial height H that you want and select the initial velocity (v0) with which it will be launched at an angle (α) with the horizontal. The gray line represents the trajectory that it will travel based on the values you have selected.

Then press the play button. Drag the time and observe as the position (x and y) and velocity (vx and vy) is calculated for each instant of its descent to the ground.

Verify that the projection on the y-axis (green) describes a vertical launch motion and on the x-axis (red) it describes a uniform rectilinear motion.

### Equation of position and trajectory in projectile motion

The equation of position of a body helps us to know at what point it is at each instant of time. In the case of a body moving in two dimensions, remember that, generically, it is described by:

$\stackrel{\to }{r}\left(t\right)=x\left(t\right)\stackrel{\to }{i}+y\left(t\right)\stackrel{\to }{j}$

Substituting the above expressions of the position on the horizontal axis (u.r.m.) and on the vertical axis (u.a.r.m.) in the generic equation of position, we can get the expression of the equation of position for parabolic motion.

The equation of position of the projectile motion is given by:

$\stackrel{\to }{r}=\left({x}_{0}+{v}_{0x}\cdot t\right)·\stackrel{\to }{i}+\left(H+{v}_{0y}·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}$

On the other hand, to know which trajectory the body follows, that is, its equation of trajectory, we can combine the above equations to eliminate t, getting:

$y=H+{v}_{0y}·\left(\frac{x}{{v}_{0x}}\right)-\frac{1}{2}·g·\left(\frac{x}{{v}_{0x}}{\right)}^{2}=H+{k}_{1}·x-{k}_{2}·{x}^{2}\phantom{\rule{0ex}{0ex}}{k}_{1}=\frac{{v}_{0y}}{{v}_{x}};\phantom{\rule{0ex}{0ex}}{k}_{2}=\frac{1}{2·{{v}_{0x}}^{2}}·g$

As expected, this is the equation of a parabola.

On the other hand, frequently in exercises, you would be asked for some of the following values.

## Maximum height

This value is reached when the velocity in the y-axis, v, is 0. Starting from the equation of velocity in the y-axis, and making vy = 0, we get the time t that it takes the body in get to this height. From that time, and from the equations of position, we can calculate the distance to the origin in the both axes, the x-axis and y-axis.

## Flight time

It is calculated for y = 0, the vertical component of the position. That is, the flight time is the time required for the height to become 0 (the projectile reaches the ground).

## Range

It is the maximum horizontal distance, from the starting point of the motion to the point in which the body hits the ground. Once the strong>flight time is obtained, simply substitute in the equation of position of the horizontal component.

## Angle of the trajectory

The angle of the trajectory in a given point is the same as the angle that the velocity vector form with the horizontal at that point. To calculate it, we get the components vx and vy and from the trigonometric definition of the tangent of an angle, we calculate α:

## Solved exercises worksheet

Here you can test what you have learned in this section.

#### Flight time, initial velocity and maximum height in parabolic motion

difficulty

A 1.95 m tall shot put athlete put the shot 25 meters away. Knowing that the path begins with an elevation of 40°, calculate:

1. The shot flight time
2. The shot initial velocity
3. The motion maximum height

#### Gooooooooaall !!! Paraaaaboliccccc!

difficulty

Minute 90 of the game... Lopera approaches the ball to make a direct free kick 40 meters from the goal, takes two steps back and kiiicks. The ball takes off at an elevation of 20°... and GOOOOALLL!!! GOOOOOOOALLL!!!! The ball goes in through the top corner at a height of 1.70 m!!!. After hearing this radio broadcast, can you answer the following questions?

a) From Lopera’s kick to scoring the goal, how long did it take? and what was the initial velocity of the ball at the moment of the kick?
b) What is the maximum height reached by the ball?
c) How fast was the ball going when it reached the goal?

#### Launching distance in parabolic motion

difficulty

We have a small device that can launch missiles with a velocity of . Determine how far we must be to hit a target if:

• the height from which the launch occurs is 1.8 m
• the target is at a height of 1.5 m

#### Elevation angle in a parabolic kick

difficulty

Determine the angle in relation to the horizontal, that you must kick a ball to the goal so that it scores barely touching the upper goal post, which is located at a height of 2.45 m, and 9 m away from the starting point. The ball is launched with a velocity of 82 km/h. Notice that the ball must be at the highest point of its trajectory to enter near the top corner of the goal.

## Formulas worksheet

Here is a full list of formulas for the section Projectile Motion. By understanding each equation, you will be able to solve any problem that you may encounter at this level.

Click on the icon   to export them to any compatible external program.

#### Equation of position x-axis

$x={v}_{x}\cdot t={v}_{0}·\mathrm{cos}\left(\alpha \right)·t$

#### Equation of position y-axis

$y=\mathrm{H}+{\mathrm{v}}_{0\mathrm{y}}·\mathrm{t}-\frac{1}{2}·g·{t}^{2}=\mathrm{H}+{\mathrm{v}}_{0}·\mathrm{sin}\left(\mathrm{\alpha }\right)·\mathrm{t}-\frac{1}{2}·g·{t}^{2}$

#### Equation of velocity x-axis

${v}_{x}={v}_{0x}={v}_{0}·\mathrm{cos}\left(\alpha \right)$

#### Equation of velocity y-axis

${v}_{y}={v}_{0y}-g\cdot t={v}_{0}·\mathrm{sin}\left(\alpha \right)-g\cdot t$

#### Equation of acceleration in the x-axis

${a}_{x}=0$

#### Equation of acceleration in the y-axis

${a}_{y}=-g$