Equations for Projectile Motion Explained

What Projectile Motion Actually Is

Projectile motion is just the path an object takes when you throw it and gravity is the only force acting on it. No engines, no friction, no one pushing. Just the object flying through the air while Earth pulls it back down.

That's it. The math behind it is straightforward once you stop overcomplicating things.

The Four Variables You Need to Know

Every projectile motion problem involves these four quantities:

• v₀ — initial velocity (speed and direction when you release the object)
• θ — launch angle (measured from horizontal)
• g — gravitational acceleration (9.8 m/s² on Earth, or 32 ft/s²)
• t — time elapsed since launch

Everything else in projectile motion is just combinations of these four.

The Core Equations

Here are the fundamental equations that govern projectile motion. Memorize them or write them down—you'll use them for every problem.

Velocity Components

Your initial velocity splits into two parts when you launch at an angle. One part moves horizontally, the other vertically.

Horizontal component: v₀ₓ = v₀ · cos(θ)

Vertical component: v₀ᵧ = v₀ · sin(θ)

If you're launching horizontally (like sliding a book off a table), θ = 0°, so v₀ᵧ = 0 and v₀ₓ = v₀.

Position Equations

These tell you where the object is at any time t:

Horizontal position: x = v₀ₓ · t = v₀ · cos(θ) · t

Vertical position: y = v₀ᵧ · t - ½gt² = v₀ · sin(θ) · t - ½gt²

The horizontal equation is simple because nothing slows you down (ignoring air resistance). The vertical equation has that -½gt² term because gravity pulls you down.

Velocity at Any Time

Horizontal velocity stays constant: vₓ = v₀ₓ = v₀ · cos(θ)

Vertical velocity changes: vᵧ = v₀ᵧ - gt = v₀ · sin(θ) - gt

At the peak of the trajectory, vᵧ = 0. That's a critical point you'll use constantly.

The Trajectory Equation

Eliminate time from the position equations and you get the path the object follows:

y = x · tan(θ) - (g · x²) / (2 · v₀² · cos²θ)

This is a parabola. Every projectile follows a parabolic path in ideal conditions.

Maximum Height

When the object stops going up and starts coming down, it has reached its peak. At that moment, vertical velocity is zero.

H = (v₀² · sin²θ) / (2g)

Key insight: maximum height depends only on the vertical component of your initial velocity. A shallow fast throw and a steep slow throw can reach the same height if their vertical components match.

Time of Flight

How long is the object in the air? That's the total time from launch to landing.

T = (2 · v₀ · sinθ) / g

Notice this is exactly twice the time to reach maximum height. Symmetry is your friend here.

For horizontal launch (θ = 0°), this formula breaks down to T = 0. Use the vertical position equation instead: set y = 0, solve for t.

Horizontal Range

The horizontal distance traveled from launch to landing:

R = (v₀² · sin2θ) / g

Maximum range occurs when sin2θ = 1, which means 2θ = 90°, so θ = 45°. That's the only time horizontal range is maximized.

45° gives maximum range. Anything else gives less. This is true only when launch and landing heights are equal.

Quick Reference Table

How to Solve Projectile Motion Problems

Here's the process. Follow it every time and you'll never get stuck.

Step 1: Break Down Initial Velocity

Split v₀ into horizontal and vertical components using trigonometry. Write down v₀ₓ and v₀ᵧ.

Step 2: Identify What You're Solving For

Are you finding time? Height? Distance? Pick the right equation from the table above.

Step 3: Solve Vertically and Horizontally Separately

Horizontal and vertical motion are independent. Solve each one using its own equations, then connect them through time.

Step 4: Use Time as the Bridge

Whatever time you calculate in the vertical equation, use that same time in the horizontal equation. Time is the same for both components.

Step 5: Check Your Work

Verify that horizontal velocity stayed constant and that vertical velocity at the peak is actually zero.

Common Mistakes

Using the wrong angle: Make sure you're measuring from horizontal. If someone says "60° above the ground," that's your angle. If they say "30° from the wall," you need to convert.

Forgetting that horizontal velocity is constant: Students often try to subtract gt from the horizontal component. Don't. Only vertical velocity changes.

Mixing up time to peak and total time: Time to peak is half of total time of flight. Many students use the wrong one and get answers twice as big or small.

Using range formula when heights aren't equal: The range equation R = v₀²sin2θ/g assumes you throw from and land at the same height. If you throw from a cliff or to a higher point, that formula doesn't apply.

Forgetting g's sign: In y = v₀ᵧt - ½gt², the negative sign is there because gravity pulls down. Keep it.

What Gets Ignored Here

Air resistance. This entire article assumes no air resistance. In real life, a thrown baseball slows down horizontally while it falls. The equations get messier and often require calculus or numerical methods.

For most introductory physics problems, air resistance is ignored. Your textbook problems will ignore it. Unless the problem explicitly tells you to account for drag, pretend it doesn't exist.

The Bottom Line

Projectile motion comes down to splitting one velocity into two components, then solving two separate one-dimensional motion problems that happen to share the same time variable. That's all.

Master the component breakdown. Know when vertical velocity equals zero. Remember that horizontal motion is constant velocity and vertical motion is accelerated motion. Everything else follows from those three facts.