Projectile Motion & Air Drag Simulator
Interactive Physics Lab: adjust initial velocity, launch angle, gravity ($g$), and air density ($\rho$) to compare ideal vacuum parabolas against real trajectories.
Physics of Projectile Motion & Aerodynamic Drag
In textbook kinematics (where air density $\rho = 0\text{ kg/m}^3$), a projectile launched with initial speed $v_0$ at angle $\theta$ traces a symmetric parabola because gravity ($m\vec{g}$) is the only force acting on it. Horizontal velocity $v_x = v_0 \cos\theta$ remains constant throughout the flight.
In a real atmosphere like Earth's ($\rho \approx 1.225\text{ kg/m}^3$ at sea level), the fluid exerts a quadratic aerodynamic drag force opposing the velocity vector: \[ F_d = \frac{1}{2} \rho \, C_d \, A \, v^2 \] On the canvas above, the dashed blue curve shows the ideal vacuum parabola, while the solid amber curve numerically integrates the equations of motion with air drag for a $1\text{ kg}$ sphere of radius $0.05\text{ m}$ ($C_d = 0.47$).
Guided experiment
Try this: Keep $v_0 = 50\text{ m/s}$ and $\theta = 45^\circ$ on Earth. In a vacuum, the projectile would travel $254.8\text{ m}$, but Earth's air density ($\rho = 1.225\text{ kg/m}^3$) shortens the range to $185.4\text{ m}$. Now click 🌕 Moon: with zero air drag and $g = 1.62\text{ m/s}^2$, the range jumps to $1543\text{ m}$!
Frequently Asked Questions
What launch angle achieves maximum range in a vacuum vs. in air?
In a vacuum ($\rho = 0$), maximum horizontal range over flat ground always occurs at $45^\circ$ because $\sin(2 \times 45^\circ) = 1$. In dense air, a slightly flatter angle ($37^\circ$ to $42^\circ$) travels farther because it spends less time losing momentum to drag.
Why do complementary angles (like 30° and 60°) land at the same spot in a vacuum?
In the vacuum range equation $R = \frac{v_0^2 \sin(2\theta)}{g}$, $\sin(2 \times 30^\circ) = \sin(60^\circ)$ equals $\sin(2 \times 60^\circ) = \sin(120^\circ)$. The $60^\circ$ shot climbs much higher and stays in the air longer, compensating for its lower horizontal speed.
How is the air-resistance trajectory calculated?
The simulator integrates Newton's second law using small time steps ($\Delta t = 0.004\text{ s}$), updating instantaneous accelerations $a_x = -(F_d / m)(v_x / v)$ and $a_y = -g - (F_d / m)(v_y / v)$ at each step.
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