# Projectile motion

Canonical: https://duckyhelper.com/learn/physics/projectile-motion/
Updated: 2026-10-01

Projectile motion is the curved path of an object that is launched and then moves under gravity alone. Split it into two separate motions. Horizontally, the velocity stays constant, because nothing pushes sideways (ignoring air resistance). Vertically, the object accelerates downward at \(g = 9.8\ \text{m/s}^2\). Time is the same for both parts, so find the time from the vertical motion, then use it for the horizontal distance.

## Key ideas

First split the launch velocity into components:

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

Then treat each direction with its own equations, taking up as positive:

$$
x = v_x t \qquad y = v_{0y}t - \tfrac{1}{2}g t^2 \qquad v_y = v_{0y} - g t
$$

- \(v_0\) is the launch speed (m/s) and \(\theta\) is the launch angle above the horizontal (degrees).
- \(v_x\) is the horizontal velocity (m/s). It never changes during the flight.
- \(v_{0y}\) is the starting vertical velocity and \(v_y\) is the vertical velocity at time \(t\) (m/s).
- \(x\) and \(y\) are the horizontal and vertical displacements from the launch point (m).
- \(g = 9.8\ \text{m/s}^2\) and \(t\) is time (s).

**The two directions side by side**

|  | Horizontal (x) | Vertical (y) |
| --- | --- | --- |
| Acceleration | 0 | \(-9.8\ \text{m/s}^2\) (down) |
| Velocity | Constant, \(v_0\cos\theta\) | Changes by 9.8 m/s every second |
| Equation for position | \(x = v_x t\) | \(y = v_{0y}t - \tfrac{1}{2}gt^2\) |
| At the highest point | Still \(v_x\) | \(v_y = 0\) |

For a launch and landing at the same height, two shortcuts follow from these equations. Time of flight is \(t = \frac{2v_0\sin\theta}{g}\), and range is \(R = \frac{v_0^2 \sin 2\theta}{g}\), which is largest at 45°.

## Worked examples

**Example 1: rolling off a table**

Problem: A ball rolls off a table 1.25 m high at 3.0 m/s. How long is it in the air, and how far from the table does it land?

1. Horizontal launch: \(v_x = 3.0\ \text{m/s}\) and \(v_{0y} = 0\). It falls 1.25 m, so \(y = -1.25\ \text{m}\).
2. Time from the vertical motion.

   $$
   -1.25 = 0 - \tfrac{1}{2}(9.8)t^2 \quad\Rightarrow\quad t = \sqrt{\frac{2(1.25\ \text{m})}{9.8\ \text{m/s}^2}} = 0.505\ \text{s}
   $$
3. Horizontal distance uses that same time.

   $$
   x = v_x t = (3.0\ \text{m/s})(0.505\ \text{s}) = 1.52\ \text{m}
   $$
4. The speed has 2 significant figures, so round both answers to 2.

Answer: In the air for 0.51 s, lands 1.5 m from the table

**Example 2: a kick at an angle**

Problem: A soccer ball is kicked from the ground at 20.0 m/s at 30.0° above the horizontal and lands on level ground. Find the time of flight, the range and the maximum height.

1. Components.

   $$
   v_x = 20.0\cos 30.0^\circ = 17.32\ \text{m/s} \qquad v_{0y} = 20.0\sin 30.0^\circ = 10.0\ \text{m/s}
   $$
2. Time of flight: it lands when \(y = 0\) again.

   $$
   t = \frac{2v_{0y}}{g} = \frac{2(10.0\ \text{m/s})}{9.8\ \text{m/s}^2} = 2.041\ \text{s}
   $$
3. Range.

   $$
   x = v_x t = (17.32\ \text{m/s})(2.041\ \text{s}) = 35.35\ \text{m}
   $$
4. Maximum height, where \(v_y = 0\).

   $$
   y_{\max} = \frac{v_{0y}^2}{2g} = \frac{(10.0\ \text{m/s})^2}{2(9.8\ \text{m/s}^2)} = 5.10\ \text{m}
   $$

Answer: Time 2.04 s, range 35.3 m, maximum height 5.10 m

**Example 3: off a cliff, with landing speed**

Problem: A stone is thrown horizontally at 12 m/s from a cliff 45 m high. How far from the base does it land, and how fast is it moving when it hits?

1. Time to fall 45 m.

   $$
   t = \sqrt{\frac{2(45\ \text{m})}{9.8\ \text{m/s}^2}} = 3.03\ \text{s}
   $$
2. Horizontal distance.

   $$
   x = (12\ \text{m/s})(3.03\ \text{s}) = 36.4\ \text{m}
   $$
3. Vertical speed at impact.

   $$
   v_y = g t = (9.8\ \text{m/s}^2)(3.03\ \text{s}) = 29.7\ \text{m/s}
   $$
4. Combine the two perpendicular parts with the Pythagorean theorem.

   $$
   v = \sqrt{v_x^2 + v_y^2} = \sqrt{(12)^2 + (29.7)^2} = 32.0\ \text{m/s}
   $$

Answer: Lands 36 m from the base at 32 m/s

## Common mistakes and how to fix them

- **Using the launch speed as the horizontal speed.** Fix: for an angled launch, \(v_x = v_0\cos\theta\), not \(v_0\).
- **Putting gravity in the horizontal equation.** Fix: \(a_x = 0\). Gravity only changes the vertical velocity.
- **Thinking speed is zero at the top.** Only \(v_y\) is zero. The ball still moves sideways at \(v_x\).
- **Using the range formula when the heights differ.** \(R = v_0^2\sin 2\theta/g\) only works when launch and landing heights match. Fix: otherwise solve the vertical equation for \(t\) first.
- **Calculator in radians.** Fix: check that \(\sin 30^\circ = 0.5\) before you start.

**Practice problems**

1. At the highest point of its path, what is a projectile's vertical velocity?
   A. 0 m/s
   B. 9.8 m/s
   C. Equal to its launch speed
   D. Equal to its horizontal velocity

   Answer: 0 m/s. At the top it has stopped going up and has not started coming down, so \(v_y = 0\). Its horizontal velocity is unchanged.

2. Ignoring air resistance, what is the horizontal acceleration of a thrown ball?
   A. 0 m/s²
   B. 9.8 m/s² forward
   C. 9.8 m/s² down
   D. It depends on the angle

   Answer: 0 m/s². No force acts sideways once the ball leaves the hand, so the horizontal velocity stays constant.

3. A ball rolls off a 0.80 m high table and lands 0.60 m from its edge. How fast was it rolling?

   Answer: 1.5 m/s. Fall time: \(t = \sqrt{2(0.80)/9.8} = 0.404\) s. Horizontal speed: \(v_x = 0.60 \div 0.404 = 1.48\) m/s, which is 1.5 m/s.

4. For a launch and landing at the same height, which angle gives the longest range?
   A. 30°
   B. 45°
   C. 60°
   D. 90°

   Answer: 45°. Range depends on \(\sin 2\theta\), which is largest (equal to 1) when \(2\theta = 90^\circ\), so \(\theta = 45^\circ\).

5. An arrow is shot horizontally at 50. m/s from 1.5 m above level ground. How far does it travel before landing?

   Answer: 28 m. Fall time: \(t = \sqrt{2(1.5)/9.8} = 0.553\) s. Distance: \(x = 50. \times 0.553 = 27.7\) m, which is 28 m to 2 significant figures.

## Frequently asked questions

### Why are horizontal and vertical motion independent?

Gravity pulls straight down, so it changes only the vertical velocity. A ball dropped and a ball thrown sideways from the same height hit the ground at the same time, because they have the same vertical motion. The thrown one just covers more ground on the way.

### Why doesn't a real ball reach the range the formula predicts?

Air resistance slows it in both directions, so real ranges are shorter and the best angle is usually less than 45°. High school and AP Physics 1 problems ignore air resistance unless they say otherwise.

### What is the order of steps for any projectile problem?

Draw the path and pick up as positive. Split the launch velocity into \(v_x\) and \(v_{0y}\). Use the vertical motion to find the time. Use that time in \(x = v_x t\). Combine components at the end only if a speed or angle is asked for.

## Sources

- [OpenStax College Physics 2e, 3.4 Projectile Motion](https://openstax.org/books/college-physics-2e/pages/3-4-projectile-motion), accessed 2026-10-01

## Related

- [The kinematics equations](https://duckyhelper.com/learn/physics/kinematics-equations/)
- [Newton's laws of motion](https://duckyhelper.com/learn/physics/newtons-laws-of-motion/)
- [Work and energy](https://duckyhelper.com/learn/physics/work-and-energy/)
- [Physics study guides](https://duckyhelper.com/learn/physics/)

## Try asking Ducky

- "Check my components on number 2. Did I use sine and cosine the right way around?"
- "My range answer is way too big. Can you look at my steps and find the problem?"
- "Give me a projectile problem where the landing is lower than the launch."

## Get DuckyHelper

Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. [Try it free in your browser](https://app.duckyhelper.com/?utm_source=duckyhelper.com&utm_medium=learn) or [Get the Mac app](https://duckyhelper.com/download/)