Projectile Motion — Equations, Range & Examples

Projectile motion is the motion of an object moving in two dimensions after launch while gravity is the only significant force. In the standard introductory model, air resistance is ignored, so horizontal acceleration is $0$ and vertical acceleration is $-g$.

That means most projectile motion problems become simpler once you split them into horizontal and vertical parts. If the projectile does not land at the same height, shortcut formulas like the usual range formula do not automatically apply.

Projectile Motion Definition And Main Idea

Start with the launch speed $v_0$ and launch angle $\theta$. Split the velocity into components:

$$v_{0x} = v_0 \cos \theta, \qquad v_{0y} = v_0 \sin \theta$$

Then treat each direction separately.

Horizontal motion:

$$x = v_{0x} t = v_0 \cos \theta \, t$$

Vertical motion, if the launch point is taken as $y = 0$:

$$y = v_{0y} t - \frac{1}{2}gt^2 = v_0 \sin \theta \, t - \frac{1}{2}gt^2$$

These equations apply to the basic model where gravity is constant and air resistance is neglected.

Projectile Motion Equations You Will Use Most

For introductory problems, these are the most useful results:

$$v_x = v_0 \cos \theta$$ $$v_y = v_0 \sin \theta - gt$$

If the projectile lands at the same height from which it was launched, the total time of flight is

$$T = \frac{2v_0 \sin \theta}{g}$$

the maximum height is

$$H = \frac{v_0^2 \sin^2 \theta}{2g}$$

and the horizontal range is

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

That range formula is not universal. It only works in the same-height case with no air resistance.

Why Projectile Motion Follows A Curved Path

The horizontal velocity stays constant in the basic model, but the vertical velocity keeps changing because gravity pulls downward every second.

So the object keeps moving forward at a steady horizontal rate while also speeding up downward. That combination creates the familiar parabolic path.

Projectile Motion Example

Suppose a ball is launched from level ground with speed $20\ \mathrm{m/s}$ at an angle of $30^\circ$. Ignore air resistance and use $g = 9.8\ \mathrm{m/s^2}$.

First split the launch velocity:

$$v_{0x} = 20 \cos 30^\circ \approx 17.32\ \mathrm{m/s}$$ $$v_{0y} = 20 \sin 30^\circ = 10\ \mathrm{m/s}$$

Because the projectile lands at the same height, the time of flight is

$$T = \frac{2(10)}{9.8} \approx 2.04\ \mathrm{s}$$

The range is then

$$R = v_{0x} T \approx 17.32 \times 2.04 \approx 35.3\ \mathrm{m}$$

You can also get the same result from the shortcut formula:

$$R = \frac{20^2 \sin 60^\circ}{9.8} \approx 35.3\ \mathrm{m}$$

The maximum height is

$$H = \frac{10^2}{2(9.8)} \approx 5.10\ \mathrm{m}$$

This is the standard workflow for projectile motion problems: resolve the launch velocity, check the height condition, then compute the quantity you need.

Common Projectile Motion Mistakes

Using the range formula in the wrong situation

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

works only when the projectile starts and lands at the same height and air resistance is ignored. If the landing height is different, you should go back to the position equations.

Mixing horizontal and vertical motion

Horizontal motion uses constant velocity in the basic model. Vertical motion uses constant acceleration $-g$. If you mix those rules together, signs and formulas start breaking quickly.

Forgetting to resolve the launch velocity

The angle does not go directly into every equation. Usually you first need

$$v_{0x} = v_0 \cos \theta, \qquad v_{0y} = v_0 \sin \theta$$

before you can solve the problem cleanly.

Assuming the vertical velocity is zero at both the top and bottom

At the highest point, the vertical velocity is zero for the basic model. At launch and landing, it usually is not. What changes is the sign and magnitude over time.

When Projectile Motion Is Used

Projectile motion appears in physics classes, ball-throw problems, launch-angle questions, simple engineering estimates, and any case where an object moves under gravity after release.

It is also a useful bridge between kinematics and forces. The motion equations describe what happens, while gravity explains why the vertical acceleration is downward.

A Simple Way To Set Up Any Projectile Motion Problem

If a problem feels messy, reduce it to two questions:

1. What is happening horizontally?
2. What is happening vertically?

That framing usually makes the setup much clearer than memorizing isolated formulas.

Try A Similar Projectile Motion Problem

Try the same launch speed with a $45^\circ$ angle and compare the range with the $30^\circ$ case. If you want a guided check, GPAI Solver can help you verify the setup before you do the arithmetic.