Why do you split projectile motion into horizontal and vertical parts?

Because the two directions are independent: gravity acts only vertically, so the horizontal velocity stays constant while the vertical velocity changes at g = 9.81 m/s². Treating them separately turns one curved motion into two simple straight-line (suvat) problems that share the same time.

What launch angle gives the maximum range?

On level ground with no air resistance, 45° gives the maximum range, because range depends on sin(2θ) and sin(90°) = 1. Complementary angles such as 30° and 60° give the same range.

How do you find the maximum height of a projectile?

At the highest point the vertical velocity is momentarily zero. Using v² = u² − 2gh with v = 0 gives H = (u sin θ)² / (2g), where u sin θ is the initial vertical component of velocity.

Does a heavier projectile fall faster?

No. Ignoring air resistance, all projectiles accelerate downward at the same g regardless of mass, so mass does not appear in the range or height equations. A cannonball and a marble launched identically follow the same path.

Projectile Motion Explained (Interactive) — Range, Height & Trajectory

A thrown ball follows a graceful curve — but the secret to projectile motionis that it isn’t one complicated motion at all. It’s two simple ones happening at once: constant-velocity motion sideways, and free-fall up and down. Split them apart and every projectile problem becomes ordinary suvat.

Two motions at once

Gravity only pulls downwards. So the vertical motion has a constant acceleration $g = 9.81\,\text{m s}^{-2}$ , while the horizontal motion has no acceleration at all — horizontal velocity never changes (ignoring air resistance). The two directions share only one thing: the clock.

Drop a ball and fire another horizontally at the same instant, and they hit the ground together. The bullet’s sideways speed does nothing to how fast it falls.

Launch one

Set the angle and speed, then launch. The gold arrow is the resultant velocity; the two thinner arrows are its constant horizontal component and its shrinking-then-growing vertical component. Watch the vertical velocity vanish at the very top — the moment of maximum height.

InteractiveProjectile motion: trajectory & velocity vectors

Loading interactive…

Set angle and speed, then Launch. Horizontal velocity stays fixed while the vertical velocity reverses — the resultant traces the parabola.

Text description ↓

An interactive projectile simulator. Sliders set the launch angle (5–85°) and launch speed. When launched, a ball follows a parabolic trajectory across a ground line. At the ball, three velocity arrows are drawn: a constant horizontal component, a vertical component that decreases to zero at the peak and then reverses, and their resultant. Dashed guides mark the range along the ground and the maximum height. Readouts show range, maximum height, time of flight, and the live horizontal velocity, vertical velocity and speed. For a launch speed u at angle θ with g = 9.81 m/s²: range R = u² sin(2θ)/g, maximum height H = (u sinθ)²/(2g), time of flight T = 2u sinθ/g.

The equations

Take “up” as positive and start at the origin. The vertical component obeys the usual suvat equations with $a=-g$ :

y = u_y t - \tfrac{1}{2}g t^2, \qquad v_y = u_y - g t.

The horizontal component has no acceleration, so it’s just distance = speed × time:

x = u_x t.

Everything else follows by eliminating $t$ between them. Because the two motions share the same time, the time of flight computed from the vertical motion is exactly the time you feed into the horizontal motion to get the range.

Range, height and the magic 45°

At the top of the flight the vertical velocity is momentarily zero. Setting $v_y = 0$ gives the time to the peak, and doubling it gives the total time of flight:

T = \frac{2u\sin\theta}{g}, \qquad H = \frac{(u\sin\theta)^2}{2g}.

Multiplying the horizontal speed by the time of flight, and using $2\sin\theta\cos\theta = \sin 2\theta$ , gives the range on level ground:

R = \frac{u^2\sin 2\theta}{g}.

Worked example

1Worked example — a ball kicked at an angle

A ball is kicked at $u = 20\,\text{m s}^{-1}$ at $30^\circ$ to the horizontal. Find its time of flight and range. Take $g = 9.81\,\text{m s}^{-2}$ .

Components: $u_x = 20\cos 30^\circ = 17.3\,\text{m s}^{-1}$ , $u_y = 20\sin 30^\circ = 10.0\,\text{m s}^{-1}$ .

T = \frac{2u_y}{g} = \frac{2(10.0)}{9.81} = 2.04\,\text{s}.

R = u_x T = 17.3 \times 2.04 = 35.3\,\text{m}.

Your turn

A stone is thrown horizontally at $15\,\text{m s}^{-1}$ from a cliff $20\,\text{m}$ high. How long is it in the air? ( $g = 9.81$ .)

Show the answer ↓

Horizontal launch means $u_y = 0$ . Vertical: $20 = \tfrac{1}{2}g t^2$ , so $t = \sqrt{40/9.81} = 2.02\,\text{s}$ . The horizontal speed is irrelevant to the fall time.

Common mistakes

Two motions at once

Drop a ball and fire another horizontally at the same instant, and they hit the ground together. The bullet’s sideways speed does nothing to how fast it falls.

Launch one

InteractiveProjectile motion: trajectory & velocity vectors

Loading interactive…

Set angle and speed, then Launch. Horizontal velocity stays fixed while the vertical velocity reverses — the resultant traces the parabola.

Text description ↓

The equations

Take “up” as positive and start at the origin. The vertical component obeys the usual suvat equations with $a=-g$ :

y = u_y t - \tfrac{1}{2}g t^2, \qquad v_y = u_y - g t.

The horizontal component has no acceleration, so it’s just distance = speed × time:

x = u_x t.

Range, height and the magic 45°

At the top of the flight the vertical velocity is momentarily zero. Setting $v_y = 0$ gives the time to the peak, and doubling it gives the total time of flight:

T = \frac{2u\sin\theta}{g}, \qquad H = \frac{(u\sin\theta)^2}{2g}.

Multiplying the horizontal speed by the time of flight, and using $2\sin\theta\cos\theta = \sin 2\theta$ , gives the range on level ground:

R = \frac{u^2\sin 2\theta}{g}.

Worked example

1Worked example — a ball kicked at an angle

A ball is kicked at $u = 20\,\text{m s}^{-1}$ at $30^\circ$ to the horizontal. Find its time of flight and range. Take $g = 9.81\,\text{m s}^{-2}$ .

Components: $u_x = 20\cos 30^\circ = 17.3\,\text{m s}^{-1}$ , $u_y = 20\sin 30^\circ = 10.0\,\text{m s}^{-1}$ .

T = \frac{2u_y}{g} = \frac{2(10.0)}{9.81} = 2.04\,\text{s}.

R = u_x T = 17.3 \times 2.04 = 35.3\,\text{m}.

Your turn

A stone is thrown horizontally at $15\,\text{m s}^{-1}$ from a cliff $20\,\text{m}$ high. How long is it in the air? ( $g = 9.81$ .)

Show the answer ↓

Horizontal launch means $u_y = 0$ . Vertical: $20 = \tfrac{1}{2}g t^2$ , so $t = \sqrt{40/9.81} = 2.02\,\text{s}$ . The horizontal speed is irrelevant to the fall time.

Projectile Motion, Launched

Two motions at once

Launch one

The equations

Range, height and the magic 45°

Worked example

Your turn

Common mistakes

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?

Projectile Motion, Launched

Two motions at once

Launch one

The equations

Range, height and the magic 45°

Worked example

Your turn

Common mistakes

Frequently asked questions

Keep exploring

Want a tutor to walk you through it?