How to Solve Projectile Motion Problems Step by Step

The method in five steps

1. Split the motion into x and y. Horizontal velocity stays constant; vertical motion is free fall.
2. List the knowns by axis, breaking an angled launch into $v_{x0} = v_0\cos\theta$ and $v_{y0} = v_0\sin\theta$.
3. Solve the vertical direction for the time of flight.
4. Multiply that time by the horizontal velocity to get the range, $R = v_{x0} t$.
5. Combine the components if the question asks for the impact speed or its direction.

The sections below apply this recipe to horizontal and angled launches.

Split the motion into x and y

Every projectile problem gets solved the same way: break the motion into a horizontal part and a vertical part, solve them separately, and connect them with time. Once the object leaves your hand (or the table, or the cannon), the only force acting on it is gravity, so the two directions behave completely differently.

• Horizontal (x): no acceleration. The horizontal velocity never changes, so $x = v_{x0} t$.
• Vertical (y): constant downward acceleration of magnitude $g = 9.8 \text{ m/s}^2$. This is ordinary free fall.

The two directions are independent. Gravity does nothing to the horizontal velocity, and the horizontal motion does nothing to the fall. The one quantity they share is time. That single fact is the whole method: almost every projectile problem is solved by finding time from the y direction, then feeding it into the x direction.

The equations, one axis at a time

Every equation here is on the AP Physics 1 equation sheet, which prints exactly three kinematics equations, and you apply them separately in each direction. See the kinematic equations guide for how to choose between them.

In the x direction, $a_x = 0$, so only one survives:

In the y direction, all three apply with $a_y = -g$ (taking up as positive):

Before touching any equation, write down what you know in each column: $v_{x0}$, $v_{y0}$, $y_0$, and the landing height. A two-column table of knowns takes 20 seconds and prevents the most common mistake, which is mixing an x quantity into a y equation.

Horizontal vs vertical motion at a glance

The two directions behave differently, and the table lines up exactly why.

Time is the one variable shared across both columns, which is why you solve the vertical column for $t$ and hand it to the horizontal column.

Horizontal launch problems

A horizontal launch problem is the easiest type: something rolls off a table or drives off a cliff moving horizontally, so $v_{y0} = 0$. The object falls exactly like a dropped object while coasting sideways at constant speed.

The recipe:

1. Find the fall time from the drop height $h$. With $v_{y0} = 0$, the position equation reduces to $h = \frac{1}{2} g t^2$, so $t = \sqrt{2h/g}$.
2. Find the range: $R = v_0 t$, where $v_0$ is the launch speed.
3. If asked for impact velocity, compute $v_y = -g t$, then combine components: speed $= \sqrt{v_x^2 + v_y^2}$.

Notice that the launch speed never appears in step 1. A ball rolling off a cliff at 2 m/s and one at 50 m/s hit the ground at the same instant; the faster one just lands farther from the base.

Angled launch problems

When the launch velocity points at an angle $\theta$ above the horizontal, first break it into components:

For a launch that lands at the same height it started (a kick from level ground), the recipe is:

1. Time of flight: the motion is symmetric, so $t = \frac{2 v_0 \sin\theta}{g}$.
2. Range: $R = v_{x0} t$.
3. Max height: at the top of the arc $v_y = 0$, so $H = \frac{(v_0 \sin\theta)^2}{2g}$.

Combining steps 1 and 2 gives the level-ground range formula $R = \frac{v_0^2 \sin(2\theta)}{g}$. That formula is not printed on the AP equation sheet and fails the moment launch and landing heights differ, so treat it as a shortcut for checking answers, not as a starting point.

How to find time of flight

Time of flight always comes from the vertical direction, and the right move depends on where the projectile lands.

• Horizontal launch from height $h$: use $h = \frac{1}{2} g t^2$, so $t = \sqrt{2h/g}$.
• Angled launch landing at launch height: use symmetry. The trip up takes exactly as long as the trip down, so $t = 2 v_{y0} / g$.
• Landing at a different height: put everything into $y = y_0 + v_{y0} t - \frac{1}{2} g t^2$ and solve the quadratic for $t$. Keep the positive root that matches the physical situation.

A useful check: vertical velocity is zero at the top of the arc, so the time to reach the peak is $v_{y0}/g$. For a level launch, the total flight time is exactly double that.

How to find the range of a projectile

Range is the horizontal distance traveled, and there is only one equation for it: $R = v_{x0} t$. The horizontal velocity never changes, so range is that constant velocity multiplied by the total time in the air.

This means you cannot find range directly. The order is forced:

1. Solve the y direction for the time of flight.
2. Multiply that time by the horizontal velocity.

Students who memorize $R = \frac{v_0^2 \sin(2\theta)}{g}$ and reach for it on every problem get burned as soon as a projectile launches from a cliff, a ramp, or a building, because the formula silently assumes level ground. The two-step method works on every geometry, and showing the time calculation is what AP free-response graders want to see anyway.

Common mistakes and how to check your work

The errors that cost the most points:

• Using the full launch speed $v_0$ in a y equation instead of the component $v_{y0} = v_0 \sin\theta$.
• Sign confusion. Pick up as positive, set $a_y = -9.8 \text{ m/s}^2$, and keep that choice for the entire problem.
• Setting $v_y = 0$ at the landing point. Vertical velocity is zero only at the peak of the arc.
• Forgetting that a dropped object and a horizontally launched object fall for the same amount of time.

To check a numeric answer, run the same inputs through the projectile motion calculator. To build intuition for how launch angle and speed trade off, spend five minutes with the projectile launcher interactive, which lets you vary both and watch the trajectory respond.