Projectile Motion Calculator (Range, Height, Time)

What the calculator above computes

The calculator above takes three inputs, launch speed $v_0$, launch angle $\theta$, and launch height $h$, and returns the projectile's time of flight, horizontal range, and maximum height. It uses $g = 9.8 \text{ m/s}^2$, the value printed on the AP Physics 1 equation sheet, and it ignores air resistance, the same assumption every AP projectile problem makes.

Try it now: enter $v_0 = 20$ m/s, $\theta = 30^\circ$, and $h = 0$. You should get a time of flight of 2.04 s, a range of 35.3 m, and a maximum height of 5.10 m. The sections below show where each of those numbers comes from, so you can reproduce them by hand using nothing but the three kinematics equations printed on the equation sheet.

Split the launch velocity into components

Projectile motion is two separate one-dimensional problems happening at the same time. Gravity pulls straight down, so the vertical velocity changes by 9.8 m/s every second while the horizontal velocity never changes. That independence is the whole trick, and it starts with breaking $v_0$ into components:

For the example above: $v_{x0} = 20\cos 30^\circ = 20 \times 0.866 = 17.3$ m/s and $v_{y0} = 20\sin 30^\circ = 20 \times 0.500 = 10.0$ m/s.

A quick sanity check that catches most setup errors: at $0^\circ$ the motion should be purely horizontal, and $\cos 0^\circ = 1$, so cosine goes with $x$. If your components come out backwards, you swapped sine and cosine. For the full setup method, see how to solve projectile motion problems.

Time of flight

For a launch from level ground ($h = 0$) the flight is symmetric: the projectile spends as long rising as it does falling. Set $v_y = 0$ at the peak in $v_y = v_{y0} + a_y t$ with $a_y = -9.8 \text{ m/s}^2$, then double the rise time:

Plugging in the example numbers: $t = \frac{2 \times 10.0 \text{ m/s}}{9.8 \text{ m/s}^2} = 2.04$ s.

If the launch height is not zero, the symmetry breaks and you need the full position equation, $y = h + v_{y0}t - \frac{1}{2}gt^2$. Setting $y = 0$ at landing gives a quadratic in $t$; the calculator above solves it and keeps the positive root. Worked example 3 below runs the same quadratic by hand so you can see every step.

Range

Range is the horizontal velocity times the time in the air. Nothing accelerates the projectile horizontally, so $v_x$ stays at $v_0\cos\theta$ for the entire flight:

For the running example: $R = 17.3 \text{ m/s} \times 2.04 \text{ s} = 35.3$ m.

You may have memorized the shortcut $R = \frac{v_0^2\sin(2\theta)}{g}$. It is correct, but only for level ground, and it is not printed on the AP equation sheet, which lists exactly three kinematics equations. Finding the time first and then multiplying by the horizontal velocity works for every launch height and matches the work free-response graders expect to see. To drill the one-dimensional equations behind each direction on their own, use the kinematics calculator.

Maximum height

At the top of the arc the vertical velocity is zero, but the horizontal velocity is still $v_{x0}$, a favorite AP trap. Use $v_y^2 = v_{y0}^2 + 2a_y(y - y_0)$ with $v_y = 0$ at the peak:

For the example: $H = 0 + \frac{(10.0 \text{ m/s})^2}{2 \times 9.8 \text{ m/s}^2} = \frac{100}{19.6} = 5.10$ m.

Notice that maximum height depends only on the vertical component of the launch velocity. Two projectiles with the same $v_0\sin\theta$ peak at exactly the same height even if their ranges are very different. Launching from a platform just shifts the whole arc up: add $h$ to the peak height, as the formula shows.

Match the tool by hand, then go practice

A calculator only helps if you can match it by hand, because nothing on the AP exam will resolve vectors for you (a scientific or graphing calculator is allowed on both sections, but the setup is still yours). Work through the three examples below, then check every answer in the tool above. If a number disagrees, you have a components error or a dropped negative sign, and finding it is the real studying.

When you want practice that feels less like arithmetic, the projectile launcher interactive lets you predict a landing point before firing and watch the trajectory drawn in real time. For the step-by-step method, including elevated and horizontal launches, read how to solve projectile motion problems, and keep the kinematic equations guide handy for the three equations everything on this page is built from.