# Projectile Motion (Physics): Definition, Equations, Problems (w/ Examples)

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Imagine you’re manning a cannon, aiming to smash down the walls of an enemy castle so your army can storm in and claim victory. If you know how fast the ball travels when it leaves the cannon, and you know how far away the walls are, what launch angle do you need to fire the cannon at to successfully hit the walls?

This is an example of a projectile motion problem, and you can solve this and many similar problems using the constant acceleration equations of kinematics and some basic algebra.

Projectile motion is how physicists describe two-dimensional motion where the only acceleration the object in question experiences is the constant downward acceleration due to gravity.

On the Earth’s surface, the constant acceleration a is equal to g = 9.8 m/s2, and an object undergoing projectile motion is in free fall with this as the only source of acceleration. In most cases, it will take the path of a parabola, so the motion will have both a horizontal and vertical component. Although it would have a (limited) effect in real life, thankfully most high school physics projectile motion problems ignore the effect of air resistance.

You can solve projectile motion problems using the value of g and some other basic information about the situation at hand, such as the initial speed of the projectile and the direction in which it travels. Learning to solve these problems is essential for passing most introductory physics classes, and it introduces you to the most important concepts and techniques you’ll need in later courses too.

## Projectile Motion Equations

The equations for projectile motion are the constant acceleration equations from kinematics, because the acceleration of gravity is the only source of acceleration that you need to consider. The four main equations you’ll need to solve any projectile motion problem are:

v=v_0+at \\ s = \bigg(\frac{v + v_0} {2}\bigg) t \\ s = v_0t + \frac{1}{2}at^2 \\ v^2 = v_0^2 + 2as

Here, v stands for speed, v0 is the initial speed, a is acceleration (which is equal to the downward acceleration of g in all projectile motion problems), s is the displacement (from the initial position) and as always you have time, t.

These equations technically are only for one dimension, and really they could be represented by vector quantities (including velocity v, initial velocity v0 and so on), but in practice you can just use these versions separately, once in the x-direction and once in the y-direction (and if you ever had a three-dimensional problem, in the z-direction too).

It’s important to remember that these are used only for constant acceleration, which makes them perfect for describing situations where the influence of gravity is the only acceleration, but unsuitable for many real-world situations where additional forces need to be considered.

For basic situations, this is all you’ll need to describe the motion of an object, but if necessary, you can incorporate other factors, such as the height from which the projectile was launched or even solve them for the highest point of the projectile on its path.

## Solving Projectile Motion Problems

Now that you’ve seen the four versions of the projectile motion formula that you’ll need to use to solve problems, you can start thinking about the strategy you use to solve a projectile motion problem.

The basic approach is to split the problem into two parts: one for the horizontal motion and one for the vertical motion. This is technically called the horizontal component and vertical component, and each has a corresponding set of quantities, such as the horizontal velocity, vertical velocity, horizontal displacement, vertical displacement and so on.

With this approach, you can use the kinematics equations, noting that time t is the same for both horizontal and vertical components, but things like the initial velocity will have different components for the initial vertical velocity and the initial horizontal velocity.

The crucial thing to understand is that for two-dimensional motion, any angle of motion can be broken down into a horizontal component and a vertical component, but when you do this there will be one horizontal version of the equation in question and one vertical version.

Neglecting the effects of air resistance massively simplifies projectile motion problems because the horizontal direction never has any acceleration in a projectile motion (free fall) problem, since the influence of gravity only acts vertically (i.e., towards the surface of the Earth).

This means that the horizontal velocity component is just a constant speed, and the motion only stops when gravity brings the projectile down to ground level. This can be used to determine the time of flight, because it’s entirely dependent on the y-direction motion and can be worked out entirely based on the vertical displacement (i.e., the time t when the vertical displacement is zero tells you time of the flight).

[insert diagrams and examples]

## Trigonometry in Projectile Motion Problems

If the problem in question gives you a launch angle and an initial velocity, you’ll need to use trigonometry to find the horizontal and vertical velocity components. Once you’ve done this, you can use the methods outlined in the previous section to actually solve the problem.

Essentially, you create a right-angled triangle with the hypotenuse inclined at the launch angle (θ) and the magnitude of the velocity as the length, and then the adjacent side is the horizontal component of the velocity and the opposite side is the vertical velocity.

Draw the right-angled triangle as directed, and you’ll see that you find the horizontal and vertical components using the trigonometric identities:

\text{sin}\; θ = \frac{\text{opposite}}{\text{hypotenuse}}

So these can be re-arranged (and with opposite = vy and adjacent = vx, i.e., the vertical velocity component and the horizontal velocity components respectively, and hypotenuse = v0, the initial speed) to give:

v_x = v_0 cos (θ) \\ v_y = v_0 sin (θ)

[insert diagram]

This is all of the trigonometry you’ll need to do to address projectile motion problems: plugging the launch angle into the equation, using the sine and cosine functions on your calculator and multiplying the result by the initial speed of the projectile.

So to go through an example of doing this, with an initial speed of 20 m/s and a launch angle of 60 degrees, the components are:

\begin{aligned} v_x &= 20 \;\text{m/s} × \cos (60) \\ &= 10 \;\text{m/s} \\ v_y &= 20 \;\text {m/s} × \sin (60) \\ &= 17.32 \;\text{m/s} \end{aligned}

## Example Projectile Motion Problem: An Exploding Firework

Imagine a firework has a fuse designed so that it explodes at the highest point of its trajectory, and it’s launched with an initial speed of 60 m/s at an angle of 70 degrees to the horizontal.

How would you work out what height h it explodes at? And what would the time from the launch be when it explodes?

This is one of many problems that involve the maximum height of a projectile, and the trick to solving these is noting that at the maximum height, the y-component of the velocity is 0 m/s for an instant. By plugging in this value for vy and choosing the most appropriate of the kinematic equations, you can tackle this and any similar problem easily.

First, looking at the kinematic equations, this one jumps out (with subscripts added to show we’re working in the vertical direction):

v_y^2 = v_{0y}^2 + 2a_ys_y

This equation is ideal because you already know the acceleration (ay = -g), the initial velocity and the launch angle (so you can work out the vertical component vy0). Since we’re looking for the value of sy (i.e., the height h) when vy = 0, we can substitute zero for the final vertical velocity component and re-arrange for sy:

0 = v_{0y}^2 + 2a_ys_y
−2a_ys_y = v_{0y}^2
s_y = \frac{−v_{0y}^2}{2a_y}

Since it makes sense to call the upwards direction y, and since the acceleration due to gravity g is directed downwards (i.e., in the -y direction), we can change ay for -g. Finally, calling sy the height h, we can write:

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

So the only thing you need to work out to solve the problem is the vertical component of the initial velocity, which you can do using the trigonometric approach from the previous section. So with the information from the question (60 m/s and 70 degrees to the horizontal launch), this gives:

\begin{aligned} v_{0y} &= 60 \;\text{m/s} × \sin (70) \\ &= 56.38 \;\text{m/s} \end{aligned}

Now you can solve for the maximum height:

\begin{aligned} h &= \frac{v_{0y}^2}{2g} \\ &= \frac{(56.38 \; \text{m/s})^2}{2 × 9.8 \;\text{m/s}^2} \\ &= 162.19 \text{m} \end{aligned}

So the firework will explode at roughly 162 meters from the ground.

## Continuing the Example: Time of Flight and Distance Traveled

After solving the basics of the projectile motion problem based purely on the vertical motion, the remainder of the problem can be solved easily. First of all, the time from the launch that the fuse explodes can be found by using one of the other constant acceleration equations. Looking at the options, the following expression:

s_y = \bigg(\frac{v_y + v_{0y}} {2}\bigg) t \\

has the time t, which is what you want to know; the displacement, which you know for the maximum point of the flight; the initial vertical velocity; and the velocity at the time of the maximum height (which we know is zero). So based on this, the equation can be re-arranged to give an expression for the time of flight:

s_y = \bigg(\frac{v_{0y}} {2}\bigg) t \\ t = \frac{2s_y}{v_{0y}}

So inserting the values and solving for t gives:

\begin{aligned} t &= \frac{2 × 162.19 \;\text{m}} {56.38 \; \text{m/s}} \\ &= 5.75 \;\text{s} \end{aligned}

So the firework will explode 5.75 seconds after launch.

Finally, you can easily determine the horizontal distance traveled based on the first equation, which (in the horizontal direction) states:

v_x = v_{0x} + a_xt

However, noting that there is no acceleration in the x-direction, this is simply:

v_x = v_{0x}

Meaning that the velocity in the x direction is the same throughout the firework’s journey. Given that v = d/t, where d is the distance travelled, it’s easy to see that d = vt, and so in this case (with sx = d):

s_x = v_{0x}t

So you can replace v0x with the trigonometric expression from earlier, input the values and solve:

\begin{aligned} s_x &= v_0 \cos (θ) t \\ &= 60 \;\text{m/s} × \cos (70) × 5.75 \;\text{s} \\ &= 118 \;\text{m} \end{aligned}

So it will travel around 118 m before the explosion.

## Additional Projectile Motion Problem: The Dud Firework

For an additional problem to work on, imagine the firework from the previous example (initial velocity of 60 m/s launched at 70 degrees to the horizontal) failed to explode at the peak of its parabola, and instead lands on the ground unexploded. Can you calculate the total time of flight in this case? How far away from the launch site in the horizontal direction will it land, or in other words, what is the range of the projectile?

This problem works in basically the same way, where the vertical components of velocity and displacement are the main things you need to consider to determine the time of flight, and from that you can determine the range. Rather than work through the solution in detail, you can solve this yourself based on the previous example.

There are formulas for the range of a projectile, which you can look up or derive from the constant acceleration equations, but this isn’t really needed because you already know the maximum height of the projectile, and from this point it’s just in free fall under the effect of gravity.

This means you can determine the time the firework takes to fall back to the ground, and then add this to the time of flight to the maximum height to determine the total flight time. From then, it’s the same process of using the constant speed in the horizontal direction alongside the time of flight to determine the range.

Show that the time of flight is 11.5 seconds, and the range is 236 m, noting that you’ll need to calculate the vertical component of the velocity at the point it hits the ground as an intermediate step.