Terminal Velocity Calculator
Free Terminal velocity Calculator for mechanics. Enter variables to compute results with formulas and detailed steps. Get results you can export or share.
Formula
Vt = sqrt(2mg / (rho x Cd x A))
Terminal velocity equals the square root of (2 times mass times gravitational acceleration) divided by (air density times drag coefficient times cross-sectional area). This is derived by setting gravitational force (mg) equal to drag force (0.5 x rho x Cd x A x V^2) and solving for velocity.
Worked Examples
Example 1: Skydiver in Belly-Down Position
Problem: An 80 kg skydiver falls in belly-down position with Cd = 0.7 and cross-sectional area = 0.7 m2 at sea level (air density = 1.225 kg/m3).
Solution: Vt = sqrt(2 x 80 x 9.81 / (1.225 x 0.7 x 0.7))\nVt = sqrt(1569.6 / 0.60025)\nVt = sqrt(2615.1)\nVt = 51.14 m/s = 184 km/h = 114 mph\nWeight = 80 x 9.81 = 784.8 N\nDrag force at terminal velocity = 784.8 N\nTime to 95% = (51.14/9.81) x 3.0 = 15.6 seconds\nDistance to 95% = approx 540 m
Result: Terminal Velocity: 51.14 m/s (184 km/h / 114 mph) | Time to 95%: ~15.6s | Distance: ~540m
Example 2: Golf Ball Free Fall
Problem: A golf ball (mass = 0.046 kg, diameter = 42.67mm, Cd = 0.25) falls from a tall building. Calculate terminal velocity at sea level.
Solution: Area = pi x (0.02134)^2 = 0.001431 m2\nVt = sqrt(2 x 0.046 x 9.81 / (1.225 x 0.25 x 0.001431))\nVt = sqrt(0.9025 / 0.000438)\nVt = sqrt(2060.5)\nVt = 45.39 m/s = 163 km/h = 101 mph\nWeight = 0.046 x 9.81 = 0.451 N\nMach = 45.39 / 343 = 0.132
Result: Terminal Velocity: 45.39 m/s (163 km/h / 101 mph) | Mach: 0.132 | Subsonic
Frequently Asked Questions
What is terminal velocity and how is it reached?
Terminal velocity is the maximum speed an object reaches when falling through a fluid (like air) when the drag force equals the gravitational force, resulting in zero net acceleration. When you first release an object, gravity accelerates it downward at 9.8 m/s per second. As speed increases, air resistance (drag force) grows proportionally to the square of velocity. At some point, drag exactly balances gravity and the object stops accelerating, falling at a constant speed. This speed is terminal velocity. A skydiver in a belly-down position reaches about 120 mph (195 km/h) after approximately 12 seconds and 450 meters of free fall. The formula is Vt = sqrt(2mg / (rho x Cd x A)), derived from setting the gravitational force equal to the drag force equation.
What factors affect terminal velocity?
Four primary factors determine terminal velocity. Mass (or weight) is directly proportional to terminal velocity: heavier objects fall faster because they need more drag force to balance gravity. The drag coefficient depends on the object shape: streamlined shapes like teardrops (Cd = 0.04) have much lower drag than flat plates (Cd = 1.28). Cross-sectional area is the projected area perpendicular to the direction of motion. A skydiver belly-down presents about 0.7 square meters, while head-down presents only 0.3 square meters, dramatically increasing speed. Air density decreases with altitude: at 30,000 feet, air density is roughly 40% of sea level, so terminal velocity is much higher at altitude. This is why Felix Baumgartner exceeded the speed of sound during his stratospheric jump, where air density was extremely low.
What is the terminal velocity of a human body?
A human skydiver has different terminal velocities depending on body position and equipment. In the standard belly-to-earth spread position, terminal velocity is approximately 120 mph (195 km/h or 54 m/s) for an average 80 kg person with a drag coefficient of about 0.7 and cross-sectional area of 0.7 square meters. In a head-down dive position, the reduced area (about 0.3 square meters) and lower drag coefficient (about 0.4) increase terminal velocity to approximately 200 mph (320 km/h). Competitive speed skydivers wearing tight suits in a head-down position can exceed 300 mph (480 km/h). With a deployed parachute (area approximately 30 square meters, Cd approximately 1.5), terminal velocity drops to about 12-18 mph (5-8 m/s), allowing safe landing. Body weight significantly affects speed, with heavier skydivers falling faster.
How does altitude and air density affect terminal velocity?
Air density decreases exponentially with altitude, and since terminal velocity is inversely proportional to the square root of density, objects fall faster at higher altitudes. At sea level, air density is 1.225 kg/m3. At 5,000 meters (16,400 feet), it drops to about 0.736 kg/m3, increasing terminal velocity by roughly 29%. At 10,000 meters (32,800 feet, typical cruising altitude), density is about 0.414 kg/m3, increasing terminal velocity by about 72%. At 39,000 meters (Felix Baumgartner jump altitude), air density is only about 0.004 kg/m3, making terminal velocity roughly 17 times higher than at sea level. This is how Baumgartner reached 833 mph (Mach 1.25) during his fall. As he descended into denser air, drag increased and he naturally decelerated without any action needed. Temperature also affects density: hot air is less dense, slightly increasing terminal velocity on warm days.
What is the physics behind reaching terminal velocity?
The approach to terminal velocity follows a hyperbolic tangent function. The exact equation for velocity as a function of time is v(t) = Vt x tanh(g x t / Vt), where Vt is terminal velocity, g is gravitational acceleration, and t is time. Initially (small t), tanh approximates its argument, so v approximately equals g x t, showing free-fall acceleration as if no drag existed. As t grows large, tanh approaches 1, and velocity asymptotically approaches Vt. The object reaches 50% of terminal velocity in about 0.55 x Vt/g seconds, 90% in about 1.47 x Vt/g seconds, and 95% in about 1.83 x Vt/g seconds. Theoretically, terminal velocity is never exactly reached, only approached asymptotically. For a skydiver with Vt = 54 m/s, reaching 95% (51 m/s) takes about 10 seconds and roughly 350 meters of fall distance. The distance equation involves the natural log of cosh, making analytical solutions possible.
How do parachutes work to reduce terminal velocity?
Parachutes reduce terminal velocity by dramatically increasing both the cross-sectional area and drag coefficient. A typical sport parachute has a canopy area of 20-30 square meters (compared to a skydiver body of 0.7 square meters) and a drag coefficient of 1.3-1.5 (compared to 0.7 for a body). Plugging these into the terminal velocity formula, the speed drops from about 54 m/s (120 mph) to about 5-7 m/s (11-16 mph), which is a survivable landing speed. Round parachutes (military style) have higher drag but no forward glide capability. Ram-air rectangular parachutes (sport skydiving) generate lift as well as drag, allowing forward flight at 20-40 mph with controlled descent rates. Drogue chutes (small stabilizing parachutes) are used to reduce speed partially from very high velocities. Parachute deployment timing is critical because opening at too high a speed can damage the canopy or cause injury from the sudden deceleration forces.