Projectile Motion

One Ball with Two Motions at Once

Projectile motion is motion along a curved path because an object moves horizontally while also moving up and down vertically. This model ignores air resistance, so the only acceleration is gravity downward.

Analyze the motion by separating its horizontal and vertical components. The horizontal component follows uniform linear motion because its velocity stays constant. The vertical component follows uniformly accelerated motion because gravity keeps changing its velocity.

x(t)=x_0+v_0\cos\theta\ t

Here, $v_0$ is the initial speed, $\theta$ is the launch angle from the horizontal, $g$ is the magnitude of gravitational acceleration, and $t$ is the time since launch. The coordinates $x_0$ and $y_0$ are the launch position. Setting both to zero gives the shorter formulas often used in examples.

The visual below keeps the ball and initial speed fixed while you compare how the launch angle changes the peak and range.

Projectile Ball Path

Change the launch angle. The ball keeps moving horizontally while gravity changes its height.

Initial speed: $v_0=22\text{ m/s}$
Time in the air: $T=3.1\text{ s}$
Range: $R=48.4\text{ m}$
Peak height: $h_{\max}=12.1\text{ m}$

Gravity Changes Only the Vertical Part

Imagine a ball leaving a ramp at an angle. Its horizontal part keeps moving forward at the same velocity, while its vertical part slows down on the way up and speeds up on the way down.

In projectile motion without air resistance, $v_x$ stays constant, but $v_y$ changes every moment.

This sentence prevents a common mistake. The object does not lose its horizontal motion at the top. Only the vertical velocity becomes zero there.

The Ball Keeps Moving Horizontally at the Peak

At the top of the path, $v_y=0$ because the ball stops for an instant in its up-and-down motion. However, $v_x$ is still present, so the ball still moves forward.

The ball does not stop at the top. After the top, gravity makes $v_y$ negative, which means the ball starts moving downward. At the same time, $v_x$ still carries the ball forward.

Part of the motion	What happens	Motion type
Horizontal	$v_x$ stays constant	Zero horizontal acceleration
Vertical while rising	Positive $v_y$ decreases to zero	Constant acceleration $-g$
Vertical while falling	$v_y$ is negative and its magnitude grows	Constant acceleration $-g$

Range Comes from Time in the Air

Horizontal range, usually written as $R$ , is the horizontal distance from launch point to landing point. Because the horizontal motion has constant velocity, range is horizontal velocity multiplied by time in the air.

R=v_x t_{\text{air}}

If two balls have the same $v_x$ , the one that stays in the air longer lands farther away. If the air time is the same, the ball with the larger $v_x$ has the larger range.

A Ball Leaves a Table Horizontally

Suppose a ball leaves the edge of a table horizontally with $v_x=4\text{ m/s}$ and the table height is $1.25\text{ m}$ . With $g=10\text{ m/s}^2$ , the fall time comes from the vertical motion:

\begin{aligned} h&=\frac{1}{2}gt^2\\ 1.25&=\frac{1}{2}(10)t^2\\ t^2&=0.25\\ t&=0.5\text{ s} \end{aligned}

The horizontal range is:

R=v_x t=4(0.5)=2\text{ m}

So the ball lands $2\text{ m}$ from the foot of the table. Notice that fall time is determined by the vertical motion, while horizontal distance is determined by the horizontal motion during that time.