Projectile Motion Fundamentals: Formulas & Physics Equations

You throw a ball. Before it lands, you can already tell someone exactly when it'll hit the ground, how high it'll rise, and how far it'll travel. Not roughly. Exactly, to the decimal.

That's four formulas: time of flight, max height, range, and the full trajectory. Most people memorize them off a formula sheet. But none of them are arbitrary. Each one falls straight out of two simple facts about gravity, and once you see where they come from, you stop needing the sheet.

Two separate problems, one shared clock

Here's the fact that makes everything else easy: gravity only pulls straight down. It has no sideways component at all.

That means the ball's horizontal motion and vertical motion don't interact. Not approximately. Not "mostly." Completely independent, the whole flight.

Horizontally, nothing pushes or pulls the ball, so its horizontal speed never changes:

$v_x = v_0 \cos\theta$

Vertically, gravity decelerates the ball on the way up and accelerates it on the way down, at a constant rate $g \approx 9.8 \text{ m/s}^2$:

$v_y(t) = v_0 \sin\theta - g t$

Every formula in this post is just these two ideas, run on the same clock. Let's build them one at a time.

The climb always takes exactly as long as the fall

So how long is the ball actually in the air? You could try to sketch the whole arc and eyeball it, but there's a shortcut hiding in the vertical story alone.

Going up, gravity slows the ball down at a steady rate. Coming down, gravity speeds it up at that exact same rate. Since the deceleration and the acceleration are identical, the time it takes to climb to the peak has to equal the time it takes to fall back down to launch height.

That's the whole idea. If we find the time to reach the peak, we've secretly found half the answer.

At the peak, the ball is moving neither up nor down, so $v_y = 0$ right there. Set the vertical velocity formula to zero and solve for $t$:

$0 = v_0 \sin\theta - g t_{up} \quad\Rightarrow\quad t_{up} = \frac{v_0 \sin\theta}{g}$

Double it, because the fall mirrors the climb, and you get the total time of flight:

$T = 2t_{up} = \frac{2 v_0 \sin\theta}{g}$

Drag the launch dial. The angle changes, the whole flight speeds up or slows down, but $t_{up}$ and $t_{down}$ never drift apart by so much as a hundredth of a second. That's not a coincidence the formula happens to produce. It's the reason the formula has a factor of 2 in it at all.

Max height is the same formula, evaluated once

We already found the one moment that matters: the peak, at $t = t_{up}$. Max height is just the vertical position formula, evaluated at that instant.

The vertical position at any time is:

$y(t) = v_0 \sin\theta \cdot t - \frac{1}{2} g t^2$

Plug in $t_{up} = v_0 \sin\theta / g$ and simplify:

$H = \frac{(v_0 \sin\theta)^2}{2g}$

No new idea required. We just asked the position formula "where are you at the one time we already solved for," and it answered.

Range is what happens when you combine the other two

Now for the interesting part: how far does the ball travel horizontally before it lands?

Horizontal distance is easy, constant speed times time: $x(t) = v_0 \cos\theta \cdot t$. We already know the total flight time is $T = 2v_0\sin\theta / g$. Landing distance is just that formula evaluated at $t = T$:

$R = v_0 \cos\theta \cdot \frac{2v_0 \sin\theta}{g} = \frac{2v_0^2 \sin\theta\cos\theta}{g}$

There's a trig identity hiding in that last expression: $2\sin\theta\cos\theta = \sin 2\theta$. Swap it in and the formula collapses to something much cleaner:

$R = \frac{v_0^2 \sin 2\theta}{g}$

Three formulas, and every one of them is either the vertical story, the horizontal story, or both stories combined at the right instant.

Watch what each formula actually marks on the curve. $T$ is the moment the path hits the ground. $H$ is the top of the arc. $R$ is where that landing point sits on the horizontal axis. None of these are separate calculations happening off to the side, they're just labels for three specific points on one continuous path.

The path doesn't need a clock at all

Sometimes you don't care when the ball is somewhere, only where its whole path goes. Can we describe the shape of the arc without mentioning time at all?

We can, by eliminating $t$ between the two position formulas. From $x(t) = v_0\cos\theta \cdot t$, solve for time: $t = x / (v_0\cos\theta)$. Substitute that into $y(t)$ and simplify:

$y = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}\,x^2$

Look at the shape of that equation. It's $y = (\text{something}) \cdot x - (\text{something else}) \cdot x^2$. That's a parabola in $x$ and $y$ directly, no time variable anywhere. The first term, $\tan\theta$, is the initial slope, how steep the ball leaves your hand. The second term is a constant that controls how sharply the path curves back down.

Raise the launch speed and watch the curve flatten out. A faster ball still follows a parabola, just a much wider, shallower one, because that curvature coefficient shrinks as $v_0^2$ grows in the denominator. Hover anywhere on the path and you'll see its slope at that exact point, which starts at $\tan\theta$ and drops steadily to negative $\tan\theta$ by the time it lands.

What these formulas quietly assume

All four formulas rest on two assumptions: the ball lands at the same height it launched from, and nothing but gravity acts on it.

Break the first assumption and the range formula stops applying directly. Throw from a cliff and the ball gets extra time to fall, so a shallower angle than 45° actually travels farther, because that fourth second of flight time is worth more than a slightly steeper launch angle.

Break the second assumption and you get air resistance, which grows with speed and eats into range in a way none of these formulas capture. A cannonball and a beach ball leaving at the same speed and angle land in very different places.

Neither assumption breaks the reasoning we used to get here, though. Gravity still only acts vertically. Horizontal and vertical motion are still independent. Every real-world correction is a term added on top of that same foundation, not a replacement for it.

Watch it happen live

Everything above was built from formulas evaluated at specific moments: the peak, the landing, or a fixed point in space. Here's what it looks like when those same formulas run continuously.

Press play, or grab the ball and drag it along its own path. $x(t)$ climbs at a constant rate the entire flight. $y(t)$ rises, peaks, and falls. $v_y(t)$ starts positive, crosses zero at the exact top of the arc, and finishes negative, by exactly as much as it started positive. Four formulas, one moving picture.

The short version

Time of flight comes from a symmetry: the climb and the fall take equal time, so $T = 2v_0\sin\theta/g$. Max height comes from evaluating the vertical position formula at that one peak instant: $H = (v_0\sin\theta)^2 / 2g$. Range comes from combining horizontal speed with the total flight time and simplifying with a trig identity: $R = v_0^2\sin2\theta/g$. And if you eliminate time entirely, the whole path collapses into a single equation, $y = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}x^2$, a parabola with no clock in it at all.

None of these are four separate facts to memorize. They're one fact, gravity pulls straight down and nowhere else, looked at four different ways.