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# Effect of Gravity on an Artillery Projectile

by Ron Kurtus (revised 12 January 2011)

The artillery in the military uses cannons or howitzers to send explosive projectiles into enemy territory. Artillery personnel determine the horizontal displacement of the target and adjust the angle of the cannon according to the known initial velocity of the projectile. It is assumed that the effect of air resistance is negligible on the projectile.

Note: Horizontal displacement is how far the projectile has traveled from the cannon along the horizontal axis. It is a vector in a specified direction. Distance is a scalar quantity that indicates how far the object traveled along its curved path.(

See Convention for Direction in Gravity Equations for more information.)

Artillery equations start with the initial velocity of the projectile, which can be divided into its vertical and horizontal components. This results in the vertical velocity component following the equations for an object projected upward and the horizontal velocity component following the simple displacement equation.

The method to calculate the horizontal displacement of the projectile is to determine the time it takes for the projectile to reach its maximum height and return to the ground and then multiply that time by the horizontal component of the velocity.

The equation for the angle required to achieve the desired displacement comes from solving the displacement equation as a function of the angle.

Questions you may have include:

- What are the velocity components of the projectile?
- How is the displacement calculated?
- What is the equation for the angle required for a given displacement?

This lesson will answer those questions. Useful tool: Units Conversion

## Velocity components

A projectile shot from a cannon starts with an initial or muzzle velocity of **−v _{i}** at an angle

**θ**(Greek letter theta) with respect to the ground. It leaves the cannon at a displacement

**h**from the muzzle to the ground. These variables are used to determine the horizontal displacement of the projectile.

Projectile leaves cannon at angle **θ** with the ground

### Components of initial velocity

The initial velocity of the projectile can be separated into its **x** and **y** components, where **v _{x}** is the initial velocity in the

**x**or horizontal direction and

**−**

**v**is the initial velocity in the

_{y}**y**or vertical direction.

Note: According to our convention for direction, the vector_{}vhas a positive value while both_{x}vand_{i}vare negative._{y}(

See Convention for Direction in Gravity Equations for more information.)

The **x** and **y** components of the projectile initial velocity are functions of the angle of the cannon:

v_{x}= −v_{i}cos(θ)

where

**v**is the initial velocity vector_{i}**v**is the component of the initial velocity in the horizontal or_{x}**x**direction

**θ**is the angle of the cannon with respect to the ground**cos(θ)**is the cosine of the angle**θ**

and

v_{y}= v_{i}sin(θ)

where

**v**is the component of the initial velocity in the vertical or_{y}**y**direction**sin(θ)**is the sine of the angle**θ**

### Components act independently

According to the rule stated in *Horizontal Motion Unaffected by Gravity*, the perpendicular velocity components act independently of each other.

This means that the vertical motion of the projectile follows the gravity equations, with an initial velocity of **v _{y}**. The horizontal motion is simply a function of

**v**and elapsed time.

_{x}### Path of projectile

The projectile will follow a parabolic path as it moves upward until it reaches its maximum vertical displacement (**y _{m}**). It then falls to the ground. At the same time, the projectile is moving in the horizontal direction at velocity

**v**.

_{x}Path of projectile fired from cannon

The projectile leaves the cannon barrel from the height **h**, which presents an small added factor in calculating the time it takes to hit the ground, as well as the displacement the projectile travels.

### Maximum vertical displacement

The maximum vertical displacement (**y _{m}**) of the projectile is:

y_{m}= −v_{y}^{2}/2g(

See Displacement Equations for Objects Projected Upward for more information.)

Substitute **v _{y} = v_{i}sin(θ)** in the equation:

y_{m}= −v_{i}^{2}sin^{2}(θ)/2g

### Weight of projectile not a factor

The weight of the projectile determines the amount of propulsion material needed for the desired initial or muzzle velocity. However weight or mass is not a factor in gravity equations for projected objects.

## Determining displacement from cannon

The vertical component of the initial velocity acts as an initial velocity of an object projected upward. You can use the *Time Equations for Objects Projected Upward* to determine the time it takes for the projectile to hit the ground.

You can then use the simple horizontal displacement equation **x = v _{x}t** to determine how far the projectile will travel to its target.

### Time to hit the ground

The projectile moves upward until it reaches a maximum vertical displacement. Then it falls to the ground. Since the barrel of the cannon is above the ground, that height should be added in.

The time it takes an object projected upward to reach the maximum vertical displacement and then fall to a displacement below the starting point is:

t = [−v_{y}− √(v_{y}^{2 }+ 2gh)]/g

where

**t**is the total time in seconds**g**is the acceleration due to gravity (9.8 m/s^{2}or 32 ft/s^{2})**h**is the displacement below the starting point in m or ft

(

See Time Equations for Objects Projected Upward for more information.)

### Horizontal displacement of the projectile

The horizontal displacement of the projectile is then:

x = v_{x}t

x = v_{x}*[−v_{y}− √(v_{y}^{2 }+ 2gh)]/g

x = −v_{x}v_{y}/g − v_{x}√(v_{y}^{2 }+ 2gh)/g

#### If height is small

However, if **h** is small compared to the horizontal displacement, you can set **h =** 0 and the equation reduces to:

x = −v_{x}v_{y}/g − v_{x}√(v_{y}^{2})/g

Simplifying and combining terms:

x = −2v_{x}v_{y}/g

#### Put in terms of initial velocity and angle

Substitute **v _{x} = −v_{i}cos(θ)** and

**v**in the horizontal displacement equation:

_{y}= v_{i}sin(θ)

x = −2[−v_{i}cos(θ)]*[v_{i}sin(θ)]/g

x = 2v_{i}^{2}sin(θ)*cos(θ)/g

Since **sin(θ)*cos(θ) = sin(2θ)/2**, the resulting horizontal displacement equation is:

x = v_{i}^{2}sin(2θ)/g

### Example

If the cannon is at an angle of **θ **= 30°, the end of the barrel is 4 ft off the ground and the initial velocity of the projectile is **v _{i}** = 640 ft/s, how far will the projectile travel?

#### Solution

Since the 4 feet adds little to the displacement, the simple equation can be used:

x = v_{i}^{2}sin(2θ)/g

Substitute values, with **g =** 32 ft/s^{2}:

x =640_{}^{2}*sin(60°)/32 ft

x =409600*0.866/32 ft

x =11084.8 ft = 2.1 miles

Note: The 4 ft height would only add 5.5 feet to the horizontal displacement, which is insignificant at over 2 miles.

### Example 2

Suppose the angle was set at 90° − 30° = 60°. What would the displacement be?

#### Solution 2

Since 2*60° = 120° and **sin(**120°**)** = 0.866, the displacement is:

x =11084.8 ft = 2.1 miles

In other words, complementary angles 30° and 60° (90° − 30°) result in the same displacement. This is true for any such complementary angles:

θand (90° −θ)

### Example 3

The displacement for the two angles is the same. What is the difference in maximum height between the two angles?

#### Solution

For **θ **= 30°:

y_{m}= −v_{i}^{2}sin^{2}(θ)/2g

y−640_{m}=^{2}*sin^{2}(30°)/2*32

y−409600*0.25/64_{m}=

y−1600 ft_{m}=

For **θ **= 60°:

y−640_{m}=^{2}*sin^{2}(60°)/64

y−409600*0.75/64_{m}=

y−4800 ft_{m}=

The greater angle results in a greater maximum vertical displacement, but the horizontal displacement is the same.

Same displacement for complementary angles

## Determining angle to hit target

Soldiers using a mortar adjust the angle of the device to hit a target at a defined displacement. Start with the displacement equation to derive the equation for the angle **θ**:

x = v_{i}^{2}sin(2θ)/g

Solve for **θ**:

sin(2θ) = gx/v_{i}^{2}

The result has two possible solutions:

θ = [arcsin(gx/v_{i}^{2})]/2

and

θ = 90° − [arcsin(gx/v_{i}^{2})]/2

which are complementary angles.

Also, **arcsin(gx/****v _{i}^{2}**

**)**is the

*arcsine*of

**gx/**

**v**. That means that

_{i}^{2}**θ**is an angle that has a sine of

**gx/**

**v**. A scientific calculator is necessary to determine the arcsine of an angle.

_{i}^{2}### Example

A target is seen at 500 m away. The muzzle velocity of the mortar is 160 m/s. What is the setting of the mortar angle?

#### Solution

Since the displacement is in meters, **g =** 9.8 m/s^{2}. **x =** 1000 m and **v _{i} =** −160 m/s.

θ = [arcsin(gx/v_{i}^{2})]/2

θ = [arcsin(9.8*1000/(−160)^{2})]/2

θ = [arcsin(0.383)]/2

θ =22.5°/2 = 11.25°

However, that angle is too shallow to be practical. Instead, use 90° − **θ**:

90° −

θ= 78.75°

This is more of a typical angle for a mortar.

## Summary

A cannon, howitzer or mortar sends projectiles a displacement away, as determined by the initial or muzzle velocity of the projectile and the angle of inclination. The velocity of the projectile can be divided into its vertical and horizontal components. These velocity components are independent of each other.

Given the initial velocity and inclination angle, the equation for the horizontal displacement traveled is:

x = v_{i}^{2}sin(2θ)/g

Two equations determine the necessary angle to hit a target at a given horizontal displacement:

θ = [arcsin(gx/v_{i}^{2})]/2

θ = 90° − [arcsin(gx/v_{i}^{2})]/2

Be methodical in your calculations

## Resources and references

### Websites

**Trajectories** - Hyperphysics

**Trajectory for Projectile Motion** - John Hopkins University - Deprtment of Physics & Astronomy

**Projectile Motion** - Wikipedia

**Projectile Motion Equations And Formulas Calculator** - AJ Design Software

**Projectile Motion, General Solution** - Id.Mind.net

**Trajectory of a projectile** - Wikipedia

**Characteristics of a Projectile's Trajectory** - Physics Classroom

### Books

**Top-rated books on Simple Gravity Science**

**Top-rated books on Advanced Gravity Physics**

## Questions and comments

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## Effect of Gravity on an Artillery Projectile