Projectile Motion: Formulas, Equations & How to Solve Problems

Last reviewed: August 2026

Table of Contents

• What Is Projectile Motion?
• Horizontal vs. Vertical Motion
• Why Horizontal Velocity Remains Constant
• Projectile Motion Variables and Velocity Components
• Projectile Motion Equations
• How to Solve Projectile Motion Problems
• How to Calculate Maximum Height
• How to Calculate Time of Flight
• How to Calculate Horizontal Range
• Same-Height vs. Elevated-Launch Projectile Motion
• Why the Trajectory Is Parabolic
• Worked Projectile Motion Problems and Numerical Examples
• Common Mistakes
• How to Check a Projectile Motion Answer
• When to Use the Projectile Motion Calculator
• Key Takeaways
• How We Checked This Guide
• Sources & References

What Is Projectile Motion?

Projectile motion describes an object moving through the air under the influence of gravity alone, after any initial launching force has already acted. Under the ideal model used throughout this guide — no air resistance, constant gravity, and the object treated as a single point — this motion is two-dimensional: it has a horizontal component and a vertical component.^[3]

The two components don't influence each other. What connects them is time — both are happening over the exact same interval, so once you know the flight time, you can find the horizontal distance traveled during that same window.

If you want to skip manual lookups, you can use our Projectile Motion Calculator to compute trajectory components, flight time, and range automatically.

Horizontal vs. Vertical Motion

Because the two directions are independent, they behave completely differently once the object is in the air.^[1][2]

This table is the foundation for everything else in this guide: whenever you're not sure which equation to use, the question is really "am I solving a horizontal problem or a vertical one?"

Why Horizontal Velocity Remains Constant

Under the ideal no-air-resistance model, there is no horizontal force acting on a projectile once it's been launched, so its horizontal acceleration is zero — and with zero acceleration, the horizontal velocity component simply doesn't change for the rest of the flight.^[3] This is really just Newton's first law in action: with no horizontal force, there's nothing to speed the object up or slow it down horizontally.

It's important to be precise about what stays constant here. It's the horizontal velocity component that's constant — not the projectile's overall speed. The vertical component keeps changing throughout the flight (it's slowed by gravity on the way up and sped up on the way down), so the object's total speed changes continuously even though its horizontal speed doesn't.

Projectile Motion Variables and Velocity Components

Before working through any equations, it helps to define the variables clearly:

The very first mathematical step in almost every projectile problem is breaking the initial velocity into its two components:

```text

v₀x = v₀ cos θ₀

v₀y = v₀ sin θ₀

```

Everything that follows treats `v₀x` as the (constant) horizontal velocity and `v₀y` as the starting point for the vertical motion, which then changes under gravity.

Projectile Motion Equations

Here is the complete set of equations used in this guide, along with exactly when each one applies.^[1][2]

The two SAME-HEIGHT ONLY conditions are the single most important thing to get right in this entire topic — using them outside those conditions produces a wrong answer, not just an imprecise one.

How to Solve Projectile Motion Problems

Most projectile problems can be worked through with the same sequence of steps:

1. Identify what's known and what you need to find. Write down `v₀`, `θ₀`, `g`, and any height information given.
2. Resolve the initial velocity into components. Calculate `v₀x` and `v₀y` using the formulas above.
3. Treat horizontal and vertical motion as two separate problems. Horizontal motion is constant-velocity; vertical motion is constant-acceleration.
4. Use time as the link between the two directions. Whatever time you find from the vertical motion also applies to the horizontal motion, and vice versa.
5. Choose the equation that matches what you know. If launch and landing heights are equal, the simplified time-of-flight and range formulas apply. If not, use the general vertical position equation instead.
6. Solve for the unknown, then recombine components if needed. (For example, combining `vₓ` and `vᵧ` to find a final speed and direction).
7. Check your units, signs, and whether the answer is physically reasonable before finalizing it.

How to Calculate Maximum Height

Maximum height is found using `h = v₀y² / (2g)`, and it gives the height above the launch point, not necessarily the absolute height above the ground. At the highest point of the trajectory, the vertical velocity momentarily equals zero — that's actually the definition of the apex. But the vertical acceleration does not become zero there; gravity continues acting at a constant `−g` throughout the entire flight, including at the exact instant the object is at its highest point.^[1] It's the velocity, not the acceleration, that's zero at the apex.

If the launch point itself is elevated by a height `y₀` above some reference level (like the ground), the object's absolute maximum height relative to that reference is `y₀ + h`, not `h` alone.

How to Calculate Time of Flight

This is the section where the same-height assumption matters most.

Case 1 — Launch and landing at the same height: use the simplified formula:

```text

T = 2v₀ sin θ₀ / g

```

This works because the vertical motion is symmetric: the time spent rising to the apex equals the time spent falling back down to the same height.

Case 2 — Launch and landing at different heights: the simplified formula above does not apply. Instead, solve the general vertical position equation for `t`:

```text

y = y₀ + v₀y·t − ½gt²

```

Substitute the known landing height (relative to the launch point) for `y` and `y₀=0`, then solve the resulting quadratic equation for `t`. This quadratic will often have two positive roots — because the projectile can pass through a given height once while rising and again while falling — so choosing the correct root for "when it lands" requires thinking about which one physically corresponds to the landing event.

How to Calculate Horizontal Range

Horizontal range is the horizontal distance between the projectile's launch position and landing position. For a projectile that launches and lands at the same height, range is calculated with:

```text

R = v₀²sin(2θ₀) / g

```

This formula comes directly from combining the constant horizontal velocity with the same-height time of flight — it is not a separate independent fact, it's a consequence of the two formulas above it.

For a fixed launch speed, equal launch and landing heights, and negligible air resistance, this expression is maximized when `sin(2θ₀) = 1`, which happens at `θ₀ = 45°`.^[1] When air resistance is significant, the angle that maximizes range can differ from 45°, since real drag depends on the object's shape, speed, and the specific conditions of flight — there is no single fixed real-world replacement for the ideal 45° result.

For a launch and landing at different heights, the simple same-height range formula does not apply. The reliable method is to solve the vertical position equation for the correct flight time (as shown in the previous section), then use the horizontal motion to find the range: `R = vₓ × t`.

Same-Height vs. Elevated-Launch Projectile Motion

This distinction deserves its own clear summary, because getting it wrong is an important and specific error to avoid.

If a problem states or implies that the launch point and landing point are at different heights — a ball thrown from a rooftop, a ball caught above where it was thrown, an object landing in a valley below the launch point — the simplified formulas don't apply, and the general method described above is needed instead.

Why the Trajectory Is Parabolic

Under the ideal model, horizontal position increases linearly with time (`x = v₀x·t`), while vertical position depends on `t²` (`y = v₀y·t − ½gt²`). Solving the horizontal equation for `t` and substituting it into the vertical equation eliminates time and produces a direct relationship between `y` and `x`, assuming launch from the origin (`x₀=0, y₀=0`):^[1]

```text

y = x·tan θ₀ − [g / (2v₀²cos²θ₀)]x²

```

This has the form `y = ax + bx²` — the equation of a parabola. This is why, under the ideal no-air-resistance model, projectile trajectories are consistently parabolic in shape.

Worked Projectile Motion Problems and Numerical Examples

Example 1: Resolving Velocity into Components

Problem: A projectile launches at 70.0 m/s at 75.0° above horizontal. Find the horizontal and vertical velocity components.

```text

v₀x = 70.0 × cos(75.0°) ≈ 18.1 m/s

v₀y = 70.0 × sin(75.0°) ≈ 67.6 m/s

```

This step is the starting point for almost every other calculation in this guide.

Example 2: Maximum Height and Time to Reach It

Problem: Using the launch from Example 1, find the maximum height and the time it takes to reach it.

```text

v₀y = 67.61 m/s (from Example 1, shown here to full precision)

h = v₀y² / (2g) = (67.61)² / (2 × 9.8) ≈ 233.3 m

t_apex = v₀y / g = 67.61 / 9.8 ≈ 6.90 s

```

The projectile reaches a maximum height of about 233 m above its launch point, 6.90 seconds after launch.

Example 3: Horizontal Distance Traveled by the Apex

Problem: For the same launch, find how far the projectile has traveled horizontally by the time it reaches its maximum height.

```text

x = v₀x × t = 18.1 × 6.90 ≈ 125 m

```

This example shows the horizontal and vertical motions being linked through the shared time value, even though each is calculated with completely separate equations.

Example 4: Elevated Landing (Different Height)

Problem: A ball is launched at 30.0 m/s at 45° above horizontal and lands 10.0 m above its launch point. Find the total time of flight and the ball's final speed and direction.

```text

Components:

v₀x = 30.0 × cos(45°) ≈ 21.2 m/s

v₀y = 30.0 × sin(45°) ≈ 21.2 m/s

Setting up the vertical position equation (y = +10.0 m):

10.0 = 21.2t − 4.9t²

4.9t² − 21.2t + 10.0 = 0

Solving this quadratic gives two positive roots: t ≈ 0.54 s and t ≈ 3.79 s. Both correspond to moments when the ball is at the 10.0 m height — once on the way up, and once on the way back down. Since the ball lands on the way down, the correct time is the later root:

t ≈ 3.79 s

Finding final velocity:

vᵧ = v₀y − gt = 21.2 − 9.8(3.79) ≈ −15.9 m/s

vₓ = 21.2 m/s (unchanged)

Final speed = √(vₓ² + vᵧ²) = √(21.2² + 15.9²) ≈ 26.5 m/s

Direction = tan⁻¹(15.9 / 21.2) ≈ 36.9° below horizontal

```

This example illustrates why the simplified same-height formulas can't be used everywhere: plugging this problem's numbers into `T = 2v₀sinθ₀/g` would give a single result that doesn't account for the landing height being different from the launch height.

Example 5: Solving for an Unknown Launch Speed

Problem: A ball needs to travel 90.0 m horizontally, launched at 30° above horizontal, landing at the same height it was launched from. What launch speed is required?

```text

R = v₀²sin(2θ₀) / g

v₀ = √(Rg / sin(2θ₀))

v₀ = √[(90.0 × 9.8) / sin(60°)]

v₀ ≈ 31.9 m/s

```

This example shows the range formula being used in reverse — solving for the launch speed needed to achieve a target distance.

Common Mistakes

• Using the total initial speed instead of its components. Vertical equations need `v₀y`, not `v₀`; horizontal equations need `v₀x`, not `v₀`.
• Mixing horizontal and vertical equations. Horizontal motion never involves `g`; vertical motion always does.
• Forgetting that `aᵧ = −g` applies for the entire flight, not just on the way up.
• Saying acceleration is zero at the apex. Only the vertical velocity is zero there — acceleration remains `−g` throughout.
• Applying the same-height range or time-of-flight formula to an elevated launch. This is an important and specific error to avoid — see Example 4 above for the correct method.
• Mixing degrees and radians inconsistently when evaluating trigonometric functions.
• Assuming 45° always gives the maximum range, without the ideal-model, same-height, fixed-speed qualifications.
• Confusing "height above the launch point" with "height above the ground" when the launch point itself is elevated.

How to Check a Projectile Motion Answer

A few quick checks catch most errors:

• Check units. Every term in an equation should resolve to the same unit (for example, every term in the vertical position equation should come out in meters).
• Check the sign of the vertical velocity. It should be positive while rising and negative while falling, given the convention used in this guide (upward = positive).
• Substitute your calculated time back into the vertical position equation and confirm it actually produces the height you expected.
• Cross-check range using `R = vₓ × t` once you already know the time — if this doesn't match a range you calculated another way, something is inconsistent.
• Compare your hand-worked answer against an interactive calculator for a final check.

These are consistency checks, not formal proof: matching results from two different methods is strong supporting evidence that a solution is correct, but it does not by itself guarantee that every step in the setup was conceptually right.

When to Use the Projectile Motion Calculator

Once you understand how these equations fit together, the Projectile Motion Calculator is a fast way to check a hand-worked answer or explore how changing the launch angle or speed affects the result. It's most useful after you've already worked through the setup yourself — using it as a first step tends to skip the part of the problem that actually builds understanding.

Key Takeaways

• Projectile motion splits into two independent problems: constant-velocity horizontal motion and constant-acceleration vertical motion, connected only by time.
• Horizontal velocity stays constant under the ideal no-air-resistance model because horizontal acceleration is zero.
• Vertical acceleration remains `−g` throughout the entire flight, including at the highest point — only the vertical velocity reaches zero there.
• The simplified time-of-flight and range formulas apply only when launch and landing heights are equal; elevated launches require solving the general position equation.
• 45° maximizes range only under ideal, same-height, fixed-speed conditions — real-world optimal angles vary depending on drag and projectile characteristics.
• Working a problem systematically — components first, then time, then recombining — helps prevent common setup errors.

How We Checked This Guide

Every equation in this guide was checked against OpenStax University Physics and OpenStax College Physics, and the conceptual explanation of constant horizontal velocity was compared against The Physics Classroom, an independent educational resource. All five worked examples, including the trajectory-equation relationship, were independently recalculated and matched the expected results. Units, sign conventions, and the same-height/elevated-launch distinction were specifically checked throughout. All 11 calculation modes and representative scenarios — including standard angled launches, elevated landings, horizontal drops, vertical launches, decimal inputs, Moon gravity, and complementary angles — were tested and verified against the site's Projectile Motion Calculator.

Sources & References

1. OpenStax — University Physics Volume 1, "4.3 Projectile Motion"

https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion

1. OpenStax — College Physics 2e, "3.4 Projectile Motion"

https://openstax.org/books/college-physics-2e/pages/3-4-projectile-motion

1. The Physics Classroom — "What is a Projectile?"

https://www.physicsclassroom.com/class/vectors/Lesson-2/What-is-a-Projectile