t 0.00 s x 0.0 m y 0.0 m |v| 0.0 m/s Earth · g = 9.81 m/s² Shots left 10Targets hit 0/5

Shot results

Range
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Time of flight
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Max height
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Impact speed
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Impact angle
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Vacuum range
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Energy now (J)
Kinetic (motion) 0Potential (height) 0Work done against air 0

Shots

No shots yet. Drag back from the launcher and let go.

1. Two motions at once

A projectile is any object that, once launched, moves under gravity (and, in reality, air). The key idea is that horizontal and vertical motion are independent: gravity acts only vertically, so in a vacuum the horizontal velocity never changes while the vertical velocity changes at a steady rate $g$. Resolve the launch velocity $v_0$ at angle $\theta$ into components:

$$v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta$$

Horizontal (no force, $a_x = 0$):

$$\begin{gathered}x(t) = v_0\cos\theta\; t\\ v_x = v_0\cos\theta\end{gathered}$$

Vertical (constant $a_y = -g$):

$$\begin{gathered}y(t) = h + v_0\sin\theta\; t - \tfrac12 g t^2\\ v_y = v_0\sin\theta - g t\end{gathered}$$

Switch on vectors in the lab: in a vacuum the blue horizontal arrows stay the same length all the way along the path, the green vertical arrows shrink to zero at the apex and grow downward, and the orange acceleration arrow is identical everywhere.

v₀cosθv₀sinθv₀ θ apex: v_y = 0g H R

2. Results for launch from level ground (vacuum)

Setting $y = 0$ (with $h=0$) gives the time of flight, and substituting into $x(t)$ gives the range. The apex is where $v_y = 0$.

$$\begin{gathered}T = \frac{2v_0\sin\theta}{g}, \qquad H = \frac{v_0^2\sin^2\theta}{2g}\\[4pt] R = \frac{v_0^2\sin 2\theta}{g}\end{gathered}$$

Because $\sin 2\theta$ peaks at $2\theta = 90^\circ$, the range is maximum at $\theta = 45^\circ$, giving $R_{max} = v_0^2/g$. Complementary angles ($30^\circ$ and $60^\circ$, say) give the same range in a vacuum, though the higher one stays in the air longer. The path is a parabola: eliminating $t$,

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

Launching from a height $h$: the projectile has extra time to travel, and the best angle falls below 45°:

$$\begin{gathered}\tan\theta_{opt} = \frac{v_0}{\sqrt{v_0^2 + 2gh}}\\[4pt] R_{max} = \frac{v_0}{g}\sqrt{v_0^2 + 2gh}\end{gathered}$$

Energy conservation gives the impact speed without solving for the path: $v_{impact} = \sqrt{v_0^2 + 2gh}$ (vacuum only).

3. Air resistance

For everyday balls at everyday speeds the Reynolds number (a measure of how turbulent the air flow is) is large (104–105), so drag is close to quadratic in speed and points opposite to the velocity of the ball relative to the air:

$$\begin{gathered}\vec F_D = -\tfrac12\,\rho\, C_d\, A\, |\vec v_{rel}|\,\vec v_{rel}\\ \vec v_{rel} = \vec v - \vec v_{wind}, \qquad A = \tfrac{\pi d^2}{4}\end{gathered}$$

Here $\rho$ is the air density (1.225 kg/m³ at sea level on Earth), $C_d$ the drag coefficient (about 0.47 for a smooth sphere, roughly 0.25 for a dimpled golf ball or a football at speed) and $A$ the frontal area. Drag couples the two directions (the horizontal deceleration depends on the total speed), so there is no closed-form solution. The lab integrates Newton's second law numerically with the fourth-order Runge–Kutta method (RK4):

$$\begin{gathered}\frac{d\vec v}{dt} = -g\,\hat y - \frac{\rho C_d A}{2m}\,|\vec v_{rel}|\,\vec v_{rel}\\ \frac{d\vec r}{dt} = \vec v\end{gathered}$$

What matters is the ratio $\rho C_d A / m$: a heavy, dense shot put (about 640 kg of mass per m² of frontal area) barely notices the air, while a ping-pong ball (about 2 kg/m²) is dominated by it.

4. Terminal velocity

Falling straight down, the drag grows until it balances the weight, $mg = \tfrac12\rho C_d A v_t^2$:

$$v_t = \sqrt{\frac{2mg}{\rho\,C_d\,A}}$$

Terminal velocity is a useful single number for how strongly air matters: if launch speeds are small compared with $v_t$, the vacuum model is good; if they are comparable or larger, drag changes the path dramatically. A ping-pong ball has $v_t \approx 8.3$ m/s, a golf ball about 45 m/s, and a cannonball about 140 m/s (see the Data tab).

5. Why the best angle is lower with drag

With drag, a high lob spends a long time in the air, losing speed the whole time, and its descent becomes steep as the horizontal velocity is eaten away (the path is no longer symmetric: it falls more steeply than it rose). A flatter launch spends less time in the air and keeps more of its horizontal speed. The optimum therefore drops below 45°, typically to 30–40° for balls, and lower still for light objects or headwinds. Try it in the challenges.

Model limits. The lab treats the projectile as a non-spinning sphere with a constant drag coefficient. Real footballs and golf balls spin, and the Magnus effect produces lift (this is why a well-struck golf drive carries far further than this model predicts) and sideways swerve. $C_d$ also changes with speed (the "drag crisis"). Air density is kept constant with height, and the ground is flat.

Gravity and air on other worlds

Worldg (m/s²)Surface air density (kg/m³)Note
Earth9.811.225Sea level, 15 °C
Moon1.620No atmosphere: always a vacuum
Mars3.71≈ 0.020Thin CO₂ atmosphere
Jupiter24.79≈ 0.16At the 1-bar cloud level (it has no solid surface)

Each example has a button that loads its numbers into the lab so you can check the answer against the simulation.

Example 1: a stone thrown on level ground (no air)

A stone is thrown at $v_0 = 20$ m/s at $\theta = 30^\circ$ on Earth ($g = 9.81$ m/s²). Find the time of flight, maximum height and range.

  1. Components: $v_{0x} = 20\cos30^\circ = 17.32$ m/s, $v_{0y} = 20\sin30^\circ = 10.00$ m/s.
  2. Time of flight: $T = 2v_{0y}/g = 2(10.00)/9.81 = 2.04$ s.
  3. Max height: $H = v_{0y}^2/(2g) = 100/19.62 = 5.10$ m, reached at $t = T/2 = 1.02$ s.
  4. Range: $R = v_{0x}T = 17.32 \times 2.039 = 35.3$ m. Check with $R = v_0^2\sin2\theta/g = 400 \times 0.866/9.81 = 35.3$ m.

Example 2: horizontal throw from a 20 m tower

A ball leaves a 20 m tower horizontally at 15 m/s. Ignoring air, when and where does it land, and how fast is it going?

  1. Vertical motion starts from rest: $20 = \tfrac12 (9.81) t^2 \Rightarrow t = \sqrt{40/9.81} = 2.02$ s.
  2. Horizontal distance: $x = 15 \times 2.019 = 30.3$ m.
  3. Vertical speed at impact: $v_y = g t = 9.81 \times 2.019 = 19.8$ m/s (downward).
  4. Impact speed $= \sqrt{15^2 + 19.8^2} = 24.8$ m/s, which agrees with energy conservation $\sqrt{v_0^2 + 2gh} = \sqrt{225 + 392.4} = 24.8$ m/s. Impact angle $= \tan^{-1}(19.8/15) = 52.9^\circ$ below horizontal.

Example 3: terminal velocity of a ping-pong ball

A ping-pong ball has $m = 2.7$ g, $d = 40$ mm and $C_d \approx 0.5$. Find its terminal velocity in air at sea level.

  1. Frontal area: $A = \pi (0.020)^2 = 1.257\times10^{-3}$ m².
  2. $v_t = \sqrt{\dfrac{2mg}{\rho C_d A}} = \sqrt{\dfrac{2(0.0027)(9.81)}{1.225 \times 0.5 \times 1.257\times10^{-3}}} = \sqrt{68.8} = 8.3$ m/s.
  3. Drop it from 50 m in the lab: the impact speed comes out at about 8.3 m/s, while in a vacuum it would be $\sqrt{2 \times 9.81 \times 50} = 31.3$ m/s.

Example 4: the same kick on the Moon

A football kicked at 25 m/s and 45° travels $R = 25^2/9.81 = 63.7$ m on Earth in a vacuum. On the Moon ($g = 1.62$ m/s², no air):

  1. $R = v_0^2/g = 625/1.62 = 386$ m, which is $9.81/1.62 = 6.06$ times further.
  2. Time of flight: $T = 2(25\sin45^\circ)/1.62 = 21.8$ s, compared with 3.6 s on Earth.

These challenges are checked automatically after every shot. A green light means you have done it.

Open experiments

  1. Complementary angles. In a vacuum, launch at 30° and then at 60° with the same speed. Compare range, height and time of flight in the shot list. Now switch air on and repeat: which one wins?
  2. Heavy versus light. Launch a shot put and a ping-pong ball at the same 30 m/s and 40°. In a vacuum they are identical. With air, measure the ratio of their ranges and connect it to their terminal velocities in the Data tab.
  3. Wind. Kick the football at 25 m/s, 40° with a 10 m/s headwind and then a 10 m/s tailwind. Is the gain from the tailwind equal to the loss from the headwind? Explain using the $|v_{rel}|\,v_{rel}$ form of drag.
  4. Energy audit. Watch the energy bar. In a vacuum KE + PE stays constant. With air, the orange "work done against air" grows. Where does that energy go?

Projectile data used in the lab

Typical values. Drag coefficients for spheres depend on speed and surface; these are representative for the speeds in the lab. Terminal velocities are for Earth sea-level air.