In a uniform gravitational field, the acceleration due to gravity, \$g\$, is constant (≈9.81 m s⁻² on Earth).
When an object is dropped from rest, its velocity increases linearly with time:
\$ v(t) = g\,t \$
and its displacement increases quadratically:
\$ s(t) = \tfrac{1}{2} g\,t^2 \$
These equations assume no other forces act on the object.
If we ignore air or liquid resistance, the motion is purely governed by gravity.
A useful analogy: imagine a ball rolling down a perfectly smooth hill—there’s nothing to slow it down, so it keeps speeding up.
When a falling object moves through a fluid (air or water), a drag force opposes its motion.
The drag force can be approximated by:
\$ Fd = \tfrac{1}{2}\,\rho\,Cd\,A\,v^2 \$
where:
This force grows with \$v^2\$, so the faster the object goes, the stronger the resistance.
The net force is then:
\$ m\,\frac{dv}{dt} = m\,g - F_d \$
which leads to a differential equation that can be solved to give a velocity that increases at first but then levels off.
Terminal velocity, \$v_t\$, occurs when the downward gravitational force equals the upward drag force:
\$ m\,g = \tfrac{1}{2}\,\rho\,Cd\,A\,vt^2 \$
Solving for \$v_t\$ gives:
\$ vt = \sqrt{\frac{2\,m\,g}{\rho\,Cd\,A}} \$
At this speed, the net force is zero, so the object falls at a constant speed.
Analogy: Think of a skydiver. Initially, the parachute is closed and the skydiver speeds up. When the parachute opens, the drag increases dramatically, and the skydiver soon reaches a steady, safe speed.
| Parameter | Symbol | Units |
|---|---|---|
| Acceleration due to gravity | \$g\$ | m s⁻² |
| Drag coefficient | \$C_d\$ | dimensionless |
| Fluid density | \$\rho\$ | kg m⁻³ |
| Cross‑sectional area | \$A\$ | m² |
| Terminal velocity | \$v_t\$ | m s⁻¹ |