Projectile motion: equations and worked examples

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?

Horizontal launch: $v_x = 3.0\ \text{m/s}$ and $v_{0y} = 0$ . It falls 1.25 m, so $y = -1.25\ \text{m}$ .
Time from the vertical motion.
$-1.25 = 0 - \tfrac{1}{2}(9.8)t^2 \quad\Rightarrow\quad t = \sqrt{\frac{2(1.25\ \text{m})}{9.8\ \text{m/s}^2}} = 0.505\ \text{s}$
Horizontal distance uses that same time.
$x = v_x t = (3.0\ \text{m/s})(0.505\ \text{s}) = 1.52\ \text{m}$
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.

Components.
$v_x = 20.0\cos 30.0^\circ = 17.32\ \text{m/s} \qquad v_{0y} = 20.0\sin 30.0^\circ = 10.0\ \text{m/s}$
Time of flight: it lands when $y = 0$ again.
$t = \frac{2v_{0y}}{g} = \frac{2(10.0\ \text{m/s})}{9.8\ \text{m/s}^2} = 2.041\ \text{s}$
Range.
$x = v_x t = (17.32\ \text{m/s})(2.041\ \text{s}) = 35.35\ \text{m}$
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?

Time to fall 45 m.
$t = \sqrt{\frac{2(45\ \text{m})}{9.8\ \text{m/s}^2}} = 3.03\ \text{s}$
Horizontal distance.
$x = (12\ \text{m/s})(3.03\ \text{s}) = 36.4\ \text{m}$
Vertical speed at impact.
$v_y = g t = (9.8\ \text{m/s}^2)(3.03\ \text{s}) = 29.7\ \text{m/s}$
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

At the highest point of its path, what is a projectile's vertical velocity?
1. 0 m/s
2. 9.8 m/s
3. Equal to its launch speed
4. Equal to its horizontal velocity
Show answer
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.
Ignoring air resistance, what is the horizontal acceleration of a thrown ball?
1. 0 m/s²
2. 9.8 m/s² forward
3. 9.8 m/s² down
4. It depends on the angle
Show answer
Answer: 0 m/s²
No force acts sideways once the ball leaves the hand, so the horizontal velocity stays constant.
A ball rolls off a 0.80 m high table and lands 0.60 m from its edge. How fast was it rolling?
Show answer
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.
For a launch and landing at the same height, which angle gives the longest range?
1. 30°
2. 45°
3. 60°
4. 90°
Show answer
Answer: 45°
Range depends on $\sin 2\theta$ , which is largest (equal to 1) when $2\theta = 90^\circ$ , so $\theta = 45^\circ$ .
An arrow is shot horizontally at 50. m/s from 1.5 m above level ground. How far does it travel before landing?
Show answer
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, accessed October 1, 2026

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

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