Projectile With and Without Air Resistance

Real projectiles don't follow perfect parabolas — air pushes back. With linear drag (a simplified but instructive model where drag is proportional to velocity), the equations of motion become:

$\dot{v}_x = -k\,v_x \qquad \dot{v}_y = -g - k\,v_y$

where $k$ is the drag coefficient (per unit mass, units of $\text{s}^{-1}$). Compare this to the vacuum case ($k = 0$) where $\dot{v}_x = 0$ — horizontal velocity stays constant. With drag, the horizontal velocity decays exponentially, which dramatically shortens the range.

The Physics

The system of ODEs has a clean closed-form solution. With initial velocity $(v_{0x}, v_{0y})$:

$v_x(t) = v_{0x}\,e^{-kt} \qquad v_y(t) = -\dfrac{g}{k} + \left(v_{0y} + \dfrac{g}{k}\right) e^{-kt}$

$x(t) = \dfrac{v_{0x}}{k}\bigl(1 - e^{-kt}\bigr)$

$y(t) = -\dfrac{g}{k}\,t + \dfrac{1}{k}\left(v_{0y} + \dfrac{g}{k}\right)\bigl(1 - e^{-kt}\bigr)$

As $t \to \infty$, $v_y \to -g/k$ — that's the terminal velocity. The visualization integrates the ODEs numerically using RK4 to get an exact pink curve regardless of the chosen $k$.

Step-by-Step Solution

Given:

• Initial speed: $v_0 = 20\;\text{m/s}$
• Launch angle: $\theta = 45°$
• Drag coefficient: $k = 0.10\;\text{s}^{-1}$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: Compare the range and trajectory with drag vs. the vacuum case.

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Step 1 — Compute the vacuum case (baseline).

$v_{0x} = v_{0y} = 20\sin 45° \approx 14.142\;\text{m/s}$

$T_{\text{vac}} = \dfrac{2 v_{0y}}{g} = \dfrac{28.284}{9.81} \approx 2.884\;\text{s}$

$R_{\text{vac}} = \dfrac{v_0^{2}}{g} = \dfrac{400}{9.81} \approx 40.77\;\text{m}$

$H_{\text{vac}} = \dfrac{v_{0y}^{2}}{2g} = \dfrac{200}{19.62} \approx 10.19\;\text{m}$

These are our reference numbers.

Step 2 — Compute the terminal velocity with drag.

$v_{\text{terminal}} = \dfrac{g}{k} = \dfrac{9.81}{0.10} = 98.1\;\text{m/s}$

That's the steady downward speed the ball would reach if it had infinite time to fall.

Step 3 — Write the position equations with drag.

Substituting $v_{0x} = v_{0y} = 14.142$, $g/k = 98.1$, and $v_{0y} + g/k = 112.242$:

$x(t) = \dfrac{14.142}{0.10}\bigl(1 - e^{-0.1t}\bigr) = 141.42\bigl(1 - e^{-0.1t}\bigr)$

$y(t) = -98.1\,t + 1122.42\bigl(1 - e^{-0.1t}\bigr)$

Step 4 — Find the landing time numerically (when $y = 0$).

The equation $-98.1\,t + 1122.42\bigl(1 - e^{-0.1t}\bigr) = 0$ is transcendental — no closed-form root. We solve it by iteration (this is exactly what the visualization does internally with RK4):

| $t$ | $e^{-0.1t}$ | $y(t)$ | |---|---|---| | 2.7 s | 0.7634 | $-264.87 + 265.56 \approx +0.69$ m | | 2.8 s | 0.7558 | $-274.68 + 274.09 \approx -0.59$ m |

Linear interpolation between these brackets:

$t_{\text{land}} \approx 2.7 + 0.1 \times \dfrac{0.69}{0.69 + 0.59} \approx 2.754\;\text{s}$

Step 5 — Compute the horizontal range with drag.

$x(2.754) = 141.42\bigl(1 - e^{-0.2754}\bigr)$

$e^{-0.2754} \approx 0.7593$

$x(2.754) = 141.42 \times (1 - 0.7593) = 141.42 \times 0.2407 \approx 34.04\;\text{m}$

Step 6 — Compute the loss.

$\text{Range lost} = R_{\text{vac}} - R_{\text{drag}} = 40.77 - 34.04 \approx 6.73\;\text{m}$

$\text{Percent loss} = \dfrac{6.73}{40.77} \times 100\% \approx 16.5\%$

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Answer: With $k = 0.10\;\text{s}^{-1}$ of linear drag, the ball lands at $\approx 34.04\;\text{m}$ instead of the vacuum distance of $\approx 40.77\;\text{m}$ — a loss of about 6.73 m, or roughly 16.5% of the vacuum range. The trajectory is also no longer symmetric: the descent is steeper than the ascent because horizontal velocity has bled away. The terminal velocity for this drag coefficient is $g/k = 98.1\;\text{m/s}$.

What Drag Does (Visually)

1. The path is no longer symmetric. Descent is steeper than ascent.
2. Range is reduced — by 16% in this example, much more for higher drag.
3. Maximum height drops slightly — vertical drag opposes upward motion all the way to the apex.
4. The trajectory loses its parabolic perfection — it's a transcendental curve, not a polynomial.

Try It

• Crank the drag coefficient from 0 toward 0.5 — see the cyan (vacuum) and pink (with-drag) trajectories diverge dramatically.
• For real baseballs, $k \approx 0.05$–$0.10\;\text{s}^{-1}$. Wiffle balls and ping-pong balls have much higher $k$ and behave very differently from a baseball.
• The HUD shows the range lost in real time. At $k = 0.5$, the projectile loses over half its range to air resistance.