Motion

Free fall and terminal velocity

Free fall and terminal velocity Core 3 min read

Free fall

Free fall is the motion of an object falling under gravity alone, with no air resistance.

Near the Earth's surface, every freely falling object accelerates downward at the same rate regardless of mass: \[ g \approx 10\,\text{m/s}^2 \] This value is given in the IGCSE exam and can be used in calculations. (More precisely, \( g = 9.81\,\text{m/s}^2 \), but 10 m/s² is the standard for IGCSE.)

Key results for free fall from rest:

After 1 s: speed = 10 m/s, distance fallen = 5 m
After 2 s: speed = 20 m/s, distance fallen = 20 m
After 3 s: speed = 30 m/s, distance fallen = 45 m

Distance fallen: \( d = \tfrac{1}{2}g t^2 = 5t^2 \) (for fall from rest).

Worked example — free fall

A stone is dropped from a cliff. Calculate (a) its speed after 3.0 s, and (b) the distance it has fallen.

(a) \( v = u + gt = 0 + (10)(3.0) = 30\,\text{m/s} \)

(b) \( s = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(10)(3.0)^2 = 5 \times 9 = 45\,\text{m} \)

Falling with air resistance

In reality, all falling objects experience air resistance (drag). The drag force depends on the object's speed — the faster it moves, the larger the drag force.

Two forces act on a falling object:

Weight (\( W = mg \)) — downward, constant.
Air resistance (drag) — upward, increases as speed increases.

The net downward force = Weight − Air resistance. As drag increases, the net force decreases, so acceleration decreases.

Terminal velocity

Terminal velocity is the constant (maximum) speed reached when the drag force becomes equal to the weight. At this point, the net force is zero and acceleration is zero, so the object falls at constant velocity.

Sequence of events for a sky-diver:

Immediately after jumping: weight > drag → net downward force → sky-diver accelerates.
As speed increases: drag increases → net force decreases → acceleration decreases (but still speeding up).
At terminal velocity: drag = weight → net force = 0 → constant speed (first terminal velocity, ~55 m/s for a typical sky-diver).
Parachute opens: drag suddenly increases dramatically (much larger than weight) → net upward force → sky-diver decelerates rapidly.
New (lower) terminal velocity: drag falls back to equal weight at a much lower speed (~7 m/s) → safe landing speed.

For "describe the motion" graph questions, describe each labelled section separately: state whether the object is accelerating/decelerating/at constant speed, and explain the forces responsible (weight vs drag). The explanation must reference the balance (or imbalance) of forces.

A common error: students say "at terminal velocity, all forces disappear." Wrong — both weight AND drag are still acting; they are just equal and opposite, giving a net force of zero. This is why the speed is constant, not why forces vanish.

Model answer: Why does a sky-diver reach terminal velocity?

When the sky-diver jumps, weight acts downward and air resistance acts upward. Initially, weight is greater than air resistance, so there is a net downward force and the sky-diver accelerates. As speed increases, air resistance increases. The net force decreases, so the acceleration decreases. Eventually, air resistance equals weight. The net force is zero and acceleration is zero, so the sky-diver falls at a constant speed — terminal velocity.