In this explainer, we will learn how to derive formulae for projectile motion and use them in problems.

Suppose a particle is projected from a flat horizontal plane at an angle of
from the horizontal with an initial velocity of
m⋅s^{−1}
and that no forces other than gravity act
upon it during its flight.

Recall that we can decompose the particle’s velocity (or position or acceleration) into horizontal and vertical components by the formulas and that we can express these components in a velocity vector , where and are unit vectors in the horizontal and vertical directions.

Recall secondly that the particle obeys the following laws of motion.

### Formula: Equations of Motion

If a particle is moving with initial velocity and constant acceleration , then its displacement at time is given by

Its velocity at time is given by

The quantities displacement and velocity can be decomposed into horizontal and vertical components.

In the case of projectile motion, the initial velocity is , there is no horizontal acceleration (i.e., ), and the only vertical acceleration is due to gravity (i.e., ). Therefore, and

Consider the flight of the particle. It has a vertical displacement of zero at two times: its time of launch and its time of landing .

We can use the laws of motion to derive a formula for the particle’s flight time, . Consider the vertical component of its displacement,

We want the time when , so

This equation has two solutions: the first solution corresponds to the time of launch. The second solution is

This is the formula for the flight time of the particle.

### Formula: Total Flight Time

Suppose a particle is projected from a flat horizontal plane at an angle of
from
the horizontal with an initial velocity of
m⋅s^{−1}. Write for its vertical acceleration due to
gravity and suppose that no forces other than gravity act upon it during its
flight. Then, the particle’s time of landing
, which is the same as its total flight time, is given by

The horizontal displacement of the particle at landing is called its
**range** .

We can substitute the formula for the total flight time into the expression for the horizontal displacement of the particle to derive a formula for as follows:

We can simplify this formula slightly using the double-angle identity to get

### Formula: Range

Suppose a particle is projected from a flat horizontal plane at an angle of
from the horizontal
with an initial velocity of
m⋅s^{−1}. Write for its vertical acceleration due to gravity
and suppose that no forces other than gravity act upon it during its flight. Then, the particle’s horizontal displacement at landing
, called its range, is given by

Suppose now we want to know the time the particle takes to reach its greatest height.

The particle is at its greatest height at the moment when it stops rising and starts falling, that is, when the vertical component of its velocity is zero. By taking in the formula for the vertical velocity, we have

This gives us the time taken to reach the greatest height. Observe that this time is half of the particle’s total flight time. This suggests that the particle’s trajectory is symmetrical about its highest point.

### Formula: Greatest Height

Suppose a particle is projected from a flat horizontal plane at an angle of
from the horizontal
with an initial velocity of
m⋅s^{−1}. Write
for its vertical acceleration due to gravity and suppose
that no forces other than gravity act upon it during its flight. Then, the time
at which the particle reaches its greatest height is given by

Finally, let us derive an equation for the particle’s trajectory, that is, an equation giving the particle’s vertical position in terms of its horizontal position . We have and

We rearrange the first of these equations to make the subject as follows:

Then, we substitute the expression for in the second equation as follows:

Now, we simplify, and apply the Pythagorean identity :

Observe that this equation is quadratic in . This tells us that the flight path of a projectile moving freely under gravity is a parabola.

### Formula: Position

Suppose a particle is projected from a flat horizontal plane at an angle of
from the horizontal
with an initial velocity of
m⋅s^{−1}. Write
for its vertical acceleration due to gravity and
suppose that no forces other than gravity act upon it during its flight. Write
and for its horizontal and vertical
position at time . Then, the particle’s vertical position
and horizontal position during flight
are related by the quadratic equation

While these formulae can be useful, it is not necessary to memorize them. It is much more important to understand how to derive them and related formulae by decomposing the projectile’s displacement, velocity, and acceleration into horizontal and vertical components and using the Newtonian laws of motion. For example, let us now derive a formula for the greatest height that the projectile reaches.

### Example 1: Deriving Equations Related to the Greatest Height of a Projectile

A projectile is launched from a flat horizontal plane at an angle of
from the horizontal. Its initial velocity is
m⋅s^{−1}, its
acceleration due to gravity is
m⋅s^{−2},
and no other forces act upon it during its flight. Given that its greatest
height is
, find the value
of the constants , , and .

### Answer

We consider the vertical component of the projectile’s displacement . We recall that the equation for vertical displacement is where is the vertical component of the projectile’s initial velocity and is the vertical component of its acceleration. We can express in terms of the initial velocity and the angle of projection as follows:

We also know that the projectile has a vertical acceleration of due to gravity. Therefore,

If we can find the time at which the particle reaches its greatest height, then we can substitute into this equation to find the greatest height.

Recall that we can derive an expression for using the observation that this time is the time at which the vertical component of the projectile’s velocity is zero. Using another equation of motion, , we get

We now substitute this time into to get and we simplify to get

Therefore, , , and .

We can use projectile motion formulae to solve problems about projectiles.

### Example 2: Finding the Starting Velocity of a Projectile given the Time of Flight and Starting Angle

A particle is projected from a horizontal plane at an angle of from the horizontal. After 15 seconds, the particle hits the ground. Find the starting velocity of the particle, taking and giving your answer to 1 decimal place.

### Answer

We can answer this question using the projectile motion formula for the total flight time, which is

Let us take a moment, however, to recall how that formula is derived.

We consider the vertical component of the projectile’s displacement . According to the equations of motion, where is the vertical component of the projectile’s initial velocity and is the vertical component of its acceleration. We can express in terms of the angle of projection as follows:

We also know that the projectile has a vertical acceleration of due to gravity. Therefore,

We are interested in the time when the projectile hits the ground, so we set this vertical displacement to zero:

This equation has two solutions, the first solution, , is the time of launch. The second solution corresponds to the projectile’s landing and gives the total flight time. We have which gives us a relationship between the flight time and the initial velocity into which we can substitute the given values: to one decimal place.

In the following example, we use projectile motion formulae to determine what angle of projection yields the greatest range.

### Example 3: Finding the Starting Angle that Results in the Greatest Range of a Projectile

A projectile is launched from a flat horizontal plane at an angle of
from the horizontal where
. Its initial
velocity is m/s,
its acceleration due to gravity is
m/s^{2},
and no other forces act upon it during its flight. Given that its
range on the horizontal plane is
m, find the value of
that maximizes .

### Answer

We are going to solve this question using the formula for a projectile’s range

Let us take a moment to recall how this formula is derived from the formula for a projectile’s total flight time.

We recall that the formula for the total flight time of a projectile is and that the horizontal component of its displacement can be expressed as

Substituting the total flight time into this expression will give us the range

Using the identity , we have

This will reach its maximum when the function is at its maximum. We recall that the sine function has a maximum of 1 when its argument is .

Therefore, reaches its maximum value when or .

If we know the coordinates of a point through which the projectile passes, then we can use the equation of its trajectory to calculate its angle of projection.

### Example 4: Finding the Possible Starting Angles of a Projectile given the Initial Velocity and a Point it Passes Through

A particle is projected from a horizontal plane at an angle of from the horizontal and with an initial velocity of 18 m/s. The particle passes through point situated 15 m horizontally and 8 m vertically from the point of launch. Find the possible values of , taking and giving your answer to 1 decimal place.

### Answer

We can solve this problem using the formula for a projectile’s trajectory

Let us first take a moment to recall how this formula is derived. Recall that according to the laws of motion, the particle’s horizontal position and vertical position are given in terms of its initial velocity , the angle with the horizontal , and the time taken as follows: and

To derive the equation of trajectory, we want to write the equation for in terms of by using substitution. Rearranging the first equation, we have which we can substitute into the second equation to get

Then, we apply the Pythagorean identity for

Notice that this is a quadratic equation in , but it is also a quadratic equation in . We are going to solve for using the coordinates of the point through which the particle passes to get

Expanding and simplifying, we have whose solutions we can find using the quadratic formula giving and . Applying , we have or to one decimal place.

We will now apply the formulae of projectile motion to a real-world problem.

### Example 5: Solving Real-World Problems with Projectile Motion Formulae

A rocket is launched vertically at a speed of 60 m/s from a point . When it reaches its maximum height, a capsule is ejected horizontally from it at a speed of 40 m/s. Find the horizontal distance from to the capsule’s landing point, taking and giving your answer to 1 decimal place.

### Answer

We decompose the motion of the capsule into its horizontal and vertical components. Since the capsule is launched horizontally when the rocket is at its maximum height, its vertical velocity at this point is 0 and it has a constant acceleration of in the negative -direction. The rocket also has vertical velocity 0 and constant acceleration at this point. Therefore, both rocket and capsule will take the same amount of time to reach the ground.

In this model, the only force acting on the rocket is that of gravity. The rocket reaches its maximum height when its *deceleration* due to
gravity reduces its initial vertical velocity to zero. Since gravity is
the only force in play, this takes the same amount of time as it takes for
the rocket to *accelerate* from zero at its highest point back to its
initial velocity *downward*. At this point, it hits the ground.

Thus, the capsule will take the same amount of time to reach the ground as the rocket, which is equal to the amount of time the rocket took to reach its maximum height.

Recall that the time taken by the rocket to reach its maximum height is given by the formula where is the initial speed and is the angle of projection. We have that and the rocket is launched vertically, so and . Therefore,

We can substitute this time into the equation of motion governing the horizontal displacement of the capsule where is the horizontal component of its initial velocity. We have that . Therefore, to one decimal place.

Let us finish by recapping a few important concepts from this explainer.

### Key Points

- We can decompose the motion of a projectile into horizontal and vertical components and apply the equations of motion and to derive formulae for the motion of a projectile.
- The total flight time of a projectile is given by where is its initial velocity, is its angle of projection, and is the acceleration due to gravity.
- The projectile’s range is given by
- The time at which the projectile reaches its greatest height is given by
- The projectile’s vertical position and horizontal position during flight are related by the quadratic equation
- We can use projectile motion formulae to answer questions about the trajectories of projectiles.