The Limit of Falling
If you drop a rock in a perfect vacuum, gravity will pull on it, and it will accelerate infinitely. It will fall at $9.8 , \text{m/s}$, then $19.6 , \text{m/s}$, then $29.4 , \text{m/s}$, continuing to speed up forever.
But we do not live in a vacuum. We live at the bottom of an ocean of air.
When a skydiver jumps out of a plane, gravity immediately pulls them downward. But as their velocity ($v$) increases, the aerodynamic drag force pushing up against them also increases (because drag squares with velocity).
Eventually, the skydiver will reach a speed where the upward force of air resistance exactly equals the downward force of their weight. When these two forces perfectly cancel each other out, the net force is zero. According to Newton's Second Law ($F=ma$), if the net force is zero, acceleration is zero.
The skydiver stops speeding up and continues to fall at a constant, maximum speed. This physical speed limit is known as Terminal Velocity.
Calculating Terminal Velocity
To calculate terminal velocity, we take the aerodynamic drag equation ($F_d = \frac{1}{2} \rho v^2 C_d A$) and set the Drag Force ($F_d$) equal to the object's Weight ($mg$). We then algebraicaly solve for velocity ($v$).
The Formula
Example: The Skydiver
Let's calculate the terminal velocity of a skydiver falling in a classic belly-to-earth spread-eagle position.
- Mass ($m$): $80 , \text{kg}$ (with gear).
- Air Density ($\rho$): $1.225 , \text{kg/m}^3$ (near sea level).
- Area ($A$): $0.7 , \text{m}^2$ (the surface area of a human facing the wind).
- Drag Coefficient ($C_d$): $1.0$ (roughly that of a flat brick).
- $v = \sqrt{\frac{2 \cdot 80 \cdot 9.80665}{1.225 \cdot 0.7 \cdot 1.0}} = \sqrt{\frac{1569}{0.8575}} = \sqrt{1829.7} = \mathbf{42.7 , \text{m/s}}$
$42.7 , \text{m/s}$ is approximately $153 , \text{km/h}$ or $95 , \text{mph}$. This is the absolute fastest this skydiver can fall in that specific posture.