**Projectile motion** is motion in **both horizontal and vertical directions** at the same time.

To get you used to the idea of projectile motion, imagine we fire a cannonball at an angle from the horizontal. The cannonball will be launched in the air and will travel some distance vertically before hitting the ground at a distance away from the cannon. The cannonball follows a **parabolic path**, as you can see in figure 1.

Figure 1. The cannonball follows a parabolic path and travels in a vertical and horizontal direction at the same time.

## Projectile motion: horizontal and vertical motion

Although projectile motion is the motion in both horizontal and vertical directions simultaneously, **both directions are independent of each other**. As a result, you can use the **linear motion equations** for horizontal and vertical motion separately when solving questions associated with projectile motion.

To understand this idea more, consider two marbles of the same size and weight. You release one marble from a specific height and throw the other horizontally from the same height. As long as you disregard wind resistance, both marbles will **hit the ground at the same time** because the **horizontal component does not influence the vertical motion of the marble**.

The distinction between the motion in the x and y directions is important because it shows us that we can use the linear equations of motions independently for both x and y directions. Let’s look at a few scenarios below to illustrate this concept further.

Be sure to check out our explanation on Linear Motion!

## Calculating projectile motion without an angle

To calculate **projectile motion without an angle**, we have to know the equations of motion, which are:

\[v = u + at\]

\[s = u \cdot t + \frac{1}{2} a \cdot t^2\]

\[v^2 = u^2 + 2 a \cdot s\]

In these equations, *v* is the final velocity measured in metres per second (m/s), *u* is the initial velocity measured in m/s, *a* is the acceleration measured in metres per second squared (m/s^{2}), *s* is the displacement measured in metres (m), and *t* is the time measured in seconds (s).

### Projectile motion without an angle equation examples

Suppose you have a body that rolls off a cliff with a velocity of 5 m/s. The body hits the ground at a distance d away from the base of a cliff that has a height of 30 m. Figure 3 shows the projectile motion without an angle, i.e. launched parallel to the horizontal. Calculate the range *d* covered by the object.

Figure 3. Projectile motion without an angle. Usama Adeel – StudySmarter Originals

**Solution**

To calculate d, the distance from the base of the cliff, we need to understand more about the motion in x and y directions.

Assuming there is no air resistance and just the gravitational force acting on the ball, the speed in the x-direction will be 5m/s until the ball hits the ground. In the y-direction, the ball has a **constant acceleration of 9.81m/s**^{2}, which is caused by the gravitational force.

But what’s the initial velocity in the y-direction?

As mentioned previously, because the motion in both x and y directions are independent of each other, the 5m/s velocity in the x-direction has no impact on the movement in the y-direction. Hence, the ball rolls off the cliff with an **initial velocity of 0m/s in the y-direction**.

The displacement horizontally will be -30m because the **downward direction is considered negative** along with the acceleration of free fall, which is -9.81 m/s^{2}.

Figure 4. The velocity in the y-direction will increase because of the acceleration in the y-direction. The velocity in the x-direction will stay constant. Usama Adeel – StudySmarter Originals

**For the x-direction:**

Initial velocity: u_{x} = 5 m/s

Distance traveled in the x-direction: d_{x} = ?

**For the y-direction:**

Initial velocity: u_{y} = 0 m/s

Displacement: s_{y} = -30 m

Acceleration due to free fall = a_{y} = -9.81 m/s^{2}

From the motion in the y-direction, we can calculate the time t because **time is the same in the x and y-direction**. By using the **second equation of motion** and plugging in the values, we get:

\[s_y = u_y \cdot t + \frac{1}{2} a \cdot t^2\]

\[-30 m = 0 \cdot t + \frac{1}{2} (9.81 m/s^2) \cdot t^2\]

\[t = 2.47 s\]

Therefore, the time taken for the ball to reach the ground from a height of 30m is 2.47s.

To calculate the distance travelled from the base of the cliff d_{x}, we use the **second equation of motion** again, but this time it will be for the movement in the x-direction.

\[d_x = u_x \cdot t + \frac{1}{2} a_x \cdot t^2\]

\[d_x =5 m/s \cdot (2.47 s) + \frac{1}{2} (0) \cdot (2.47s)^2\]

\[d_x = 12.35 m\]

The distance the ball travels in the x-direction with an initial velocity of 5m/s from a height of 30m is 12.35m.

You can **use any equation of motion** to calculate a specific entity depending on the problem related to a projectile motion.

## Calculating projectile motion at an angle

Above, we discussed the projectile motion of an object launched without an angle. For **projectile motion at an angle**, the **principle is the same** as projectile motion without an angle. But to make it a bit trickier, let’s solve a problem for launching and landing on different elevations.

### Projectile motion with different elevations equation examples

Take a look at the figure below. A cannonball is fired off a cliff at an initial velocity of 90m/s from a height of 25m from the ground at an angle of 53°. Calculate the distance the cannonball travels in the x-direction.

Figure 5. Projectile motion at an angle. Usama Adeel – StudySmarter Originals

**Solution**

As you can see in the figure above, the ground is elevated 9m from the base of the cliff where the cannonball will land. This means that the displacement in the y-direction will not be 25m – it will be different.

First, let us resolve the velocity vector into its components.

Figure 6. Resolution of the vector into its components. Usama Adeel – StudySmarter Originals

Initial speed in the x-direction: V_{x}=90cos53m/s

Initial speed in the y-direction: V_{y}=90sin53m/s

The displacement will be -16m as the direction downward is taken as negative.

Figure 7. Projectile motion at an angle. The displacement will be different as the ground is elevated where the ball lands. Usama Adeel – StudySmarter Originals

Using the **second equation of motion** and plugging in the values, we can calculate the time t it takes the cannonball to hit the ground from launch. Note that the time taken will be the same in either x- or y-direction.

\[s_y = u_y \cdot t + \frac{1}{2} a_y \cdot t^2\]

\[-16 = (90 \sin 53 m/s) \cdot t + \frac{1}{2} (-9.81 m/s^2) \cdot t^2\]

\[t = 14.45 s\]

As the air resistance is negligible, the **speed in the x-direction will be consistent**, i.e. it will be 90cos53 throughout the motion. We can figure out the distance travelled by multiplying speed in the x-direction with the time taken. Therefore,

\[d_x = (90 \cos 53 m/s) \cdot 14.45 s\]

\[d_x =

So, the horizontal distance the fired cannonball travels is 782.66m.

## Factors affecting projectile motion

In the two scenarios above, we assumed the air resistance was negligible. In practice, however, we can’t ignore air resistance. Similarly, various other factors influence the trajectory of a projectile motion. Let’s take a look at these factors.

### Gravity

While gravity might not directly affect the horizontal motion, the **falling time of the object will decrease if the gravity is larger**. Hence, the vertical component of the projectile motion will be limited. In turn, the object will be in the air for a lesser time, and less distance will be covered in the x-direction.

### Air resistance

The **air resistance will decrease the horizontal component **of the trajectory. However, air resistance is also influenced by other factors, such as:

- Surface-to-volume ratio: an object with a larger surface area is affected by more air resistance.
- The surface of the object: a rough surface will be impacted more by air resistance.
- Speed: if an object’s speed increases, the air resistance will also increase.

Air resistance will always affect the projectile regardless of which angle or height it is launched from.

**Speed of release** is another factor that affects projectile motion. If the launch speed is greater, the distance covered by an object will be greater as well.

### The angle of release

Considering that air resistance is negligible and both launching and landing points are at the same height, the **optimum angle for a maximum trajectory is considered to be 45°**.

However, if the release angle is more or less than 45°, then a shorter distance will be covered in either the horizontal or vertical axis. Figure 8 illustrates an object thrown at different angles and the distance covered.

In the graph, the launch speed is 10m/s, and it is assumed that there is no air resistance. T is the time of flight, t is the time from launch, R is the range, and H is the highest point of the trajectory. The length depicts the speed at each instance on the graph.

Figure 8. Trajectories of projectiles launched at different elevation angles. Cmglee CC BY-SA 3.0

### Height of release

If an object is thrown from higher up, the **distance covered will be greater** because the horizontal component will influence an object for a longer period.

Suppose you have two objects at different heights, and you want to throw each of them in such a way that both objects should cover the same distance. There is no air resistance, both objects are of the same size and mass, and both are launched at the same speed.

The object thrown from a greater height will need to be launched at a decreasing angle and vice versa for the object thrown from a lower height. Hence, there exists a relationship between the height of release and the distance covered horizontally.

### Spin

The amount of spin on a ball will also determine how much distance it can travel once you hit it. In short, the range of distance increases if you hit a backspin. The opposite occurs for topspin.

## Projectile Motion - Key takeaways

- Projectile motion is a motion of an object in a curved (parabola) path under the influence of gravity.
- Projectile motion occurs when an object moves in two dimensions, i.e. an object travels in the horizontal and vertical directions simultaneously.
- Both horizontal and vertical directions in a projectile motion are independent of each other, but their time periods will be the same.
- We can split the projectile vector into its x and y components. In a projectile motion, neither horizontal nor vertical components impact one another.
- We can also use the linear equations of motions independently for horizontal and vertical directions.

**Images**

Compound motion. https://commons.wikimedia.org/wiki/File:Compound_Motion.gif

Ideal projectile motion for different angles. https://commons.wikimedia.org/wiki/File:Ideal_projectile_motion_for_different_angles.svg

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##### Frequently Asked Questions about Projectile Motion

Does air resistance affect the motion of a projectile?

Yes, air resistance affects the motion of a projectile. Air resistance will affect the horizontal component of the projectile motion.

Does the launch angle affect the horizontal projectile motion?

Yes, the launch angle affects the horizontal projectile motion. The angle at which an object is launched does play a part in determining the distance an object travels.

Does mass affect projectile motion?

The vertical motion depends on the acceleration of free fall, whereas the horizontal motion depends on the horizontal velocity and time of flight. So, the mass does not affect projectile motion.

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