Example 1
m = 80 kg, S = 0.5 m², C = 1, blank ρ -> v ≈ 50.61 m/s.
Enter mass, area and the drag coefficient. You can leave air density blank: the calculator uses 1.225 kg/m³. Eighty kilograms, 0.5 m² and C = 1 give about 50.6 m/s.
Fall with no drag: h = ½gt². Density: ρ = m/V. Momentum: p = m·v.
Mass m (kg), Area S (m²), Coefficient C and Density ρ (kg/m³, blank = 1.225). The result shows up here.
Terminal speed is where weight balances quadratic drag: mg = ½ ρ C S v², so v = √(2mg/(ρ C S)). The calculator keeps g at 9.81 m/s². When you leave ρ blank, we insert air at the ground, 1.225 kg/m³. That empty field is useful: most school problems want that air anyway.
A person of 80 kg, S = 0.5 m², C = 1 and blank ρ: v ≈ 50.6 m/s, about 182 km/h. With a 25 m² canopy and C = 1.4 you get about 6.05 m/s, already a parachute landing speed. A 1 kg ball, S = 0.01 m², C = 0.47: about 58.4 m/s.
Larger S or C pulls v down. Twice the area at the same mass cuts v by √2, not by half. Twice the mass at the same S raises v by √2. Water ρ, about 1000, is hundreds of times air: the same card gives another scale in water.
Free fall h = ½ g t² does not know air. Speed there grows without a ceiling. Here there is a ceiling. Momentum p = m v is a different quantity: 80 kg at 50.6 m/s is about 4050 kg·m/s, but you need this v first.
m, S and C must be positive. Typed 0 in C rejects the drag model. If you want a density other than air, type it on purpose, for example 1.2 or 1000. Leave ρ blank only when you really want 1.225.
Type m, S and C, leave ρ blank or give your own, then Calculate. The example 80, 0.5, 1 and blank ρ should come back near 50.61 m/s. A comma in 0.5 parses.
v = √(2mg / (ρCS))
m > 0, S > 0, C > 0, ρ > 0. g = 9.81 m/s². Blank ρ = 1.225 kg/m³.
This calculator finds terminal speed, where weight balances quadratic drag. 80 kg, 0.5 m² and C = 1 in 1.225 kg/m³ air is about 50.6 m/s.
m = 80 kg, S = 0.5 m², C = 1, blank ρ -> v ≈ 50.61 m/s.
m = 80 kg, S = 25 m², C = 1.4, blank ρ -> v ≈ 6.05 m/s.
m = 1 kg, S = 0.01 m², C = 0.47, blank ρ -> v ≈ 58.38 m/s.
m = 0.01 kg, S = 0.001 m², C = 0.5, blank ρ -> v ≈ 17.90 m/s.
m = 0.005 kg, S = 0.06 m², C = 1.2, blank ρ -> v ≈ 1.05 m/s.
m = 7 kg, S = 0.038 m², C = 0.47, blank ρ -> v ≈ 79.24 m/s.
m = 80 kg, S = 0.5 m², C = 1, ρ = 1 kg/m³ -> v ≈ 56.03 m/s.
m = 0.145 kg, S = 0.0043 m², C = 0.3, blank ρ -> v ≈ 42.43 m/s.
v ≈ 50.6 m/s, about 182 km/h. Blank ρ is 1.225 kg/m³, air at the ground.
v ≈ 6.05 m/s. Large S pulls terminal speed down to a parachute landing.
v ≈ 58.4 m/s with blank ρ. Small ball, small area.
Because most problems want air at 1.225 kg/m³. The empty field is a shortcut here, not missing data.
There is no air there, and v grows without end. Here there is a terminal speed, when drag matches weight.
No. Momentum of 80 kg at 50.6 m/s is about 4050 kg·m/s. Here you compute v; momentum is on its own card.
That is the water scale. v drops sharply, because density is hundreds of times 1.225.
v falls as 1/√S. Twice S at fixed m gives v / √2, not half.
The calculator keeps g = 9.81. Another planet is not in the fields.
Yes. 0,5 and 0.5 are the same S. A comma and a period mean the same value.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.