9478 / 2027
Projectile Motion overview

Lesson 4 of 4

Falling with air resistance

  • I can explain falling with air resistance in a uniform gravitational field using forces and energy, including terminal velocity.Syllabus

    Syllabus 9478, 5(e). Explain falling with air resistance in a uniform gravitational field using forces and energy, including terminal velocity.

Make a guess. You are not marked.

A skydiver falling at terminal velocity opens the parachute. What happens immediately afterwards?

  1. The skydiver moves upwards, because the parachute pulls up hard.
  2. The skydiver decelerates, because drag is now greater than weight.
  3. Nothing changes at first, because the speed is already terminal.
Show the answer

The skydiver decelerates, because drag is now greater than weight.

The open parachute greatly increases drag at the same speed. The resultant force is upwards, so the skydiver slows until drag falls back to equal the weight at a lower terminal velocity.

A falling body's weight can stay constant while its acceleration changes. As drag grows, the resultant force decreases; at terminal speed the body keeps moving while the forces balance.

Consider a body released from rest in still air. Use a uniform gravitational field, constant mass and negligible buoyancy. Its shape and the surrounding air conditions remain unchanged, and drag increases with its speed relative to the air.

For this fall, choose downward as positive. This differs from the upward-positive convention used for the projectile examples. Let D be the upward drag magnitude:

Fresultant,down = mg - D
adown = g - D/m

Unchanged weight, increasing upward drag

One body is released from rest in still air. Weight is constant and buoyancy is neglected. Purple arrows are forces on the body; the separate blue arrow shows velocity.

1. Just released from rest

1. Just released from restA body is released from rest in still air. Its downward weight arrow is 120 drawing units long, the same as in the other two stages. At zero relative air speed, drag is zero, so no upward force arrow is drawn. Downward acceleration is g. Buoyancy is neglected in this model.WeightDrag = 0v = 0Same body; same downward weight

Acceleration is g downward.

2. Falling faster, still speeding up

2. Falling faster, still speeding upThe body moves downward. Weight is unchanged, with the same 120-unit downward arrow. Upward drag is smaller than weight, shown schematically by a shorter 60-unit arrow; this is not an assigned numerical force or an exact sampled instant. Resultant force and acceleration remain downward, but are smaller than at release. The separate blue arrow indicates downward velocity, not an additional force.WeightDrag < weightv downSame body; same downward weight

Downward acceleration is less than g.

3. Terminal speed

3. Terminal speedThe body continues downward at terminal speed. The upward drag and downward weight arrows are both 120 drawing units long, so the resultant force and acceleration are zero. The separate blue downward arrow shows nonzero velocity. Constant kinetic energy does not mean energy transfer has stopped: gravitational potential energy continues to decrease and energy is transferred to internal stores of the body and air.WeightDrag = weightv = vtSame body; same downward weight

Resultant force = 0; acceleration = 0.

All weight arrows have the same length; the terminal drag arrow has that same length. The middle drag arrow only means smaller than weight, not a specified fraction. Velocity arrows have different units and are not on the force scale.

At terminal speed, kinetic energy stays constant, but gravitational potential energy still decreases and energy continues to enter internal stores of the body and air.

Each force diagram refers to the same falling body. Weight stays the same while upward drag increases from zero to a value equal to weight.
  1. Just released: relative air speed is zero, so drag is zero in this model. The downward resultant is mg and downward acceleration is g.
  2. Speed increasing: drag acts upward and grows. It is smaller than weight, so the body still accelerates downward, but its acceleration is less than g.
  3. Terminal motion: drag balances weight. The resultant and acceleration are zero while the body continues downward at a nonzero constant speed.

For these falling graphs, downward is positive

This convention differs from the upward-positive launch graphs. The same body falls in unchanged still-air conditions. These are qualitative trends, with no numerical time scale or assumed drag formula.

Downward speed approaches a limit

Downward speed approaches a limitA schematic graph starts at zero downward speed at release and rises with decreasing positive gradient towards terminal speed v subscript t. It stays below the dashed limiting value, with no overshoot. Time has no numerical scale and the curve does not assign an exact time to reach terminal speed or a specific drag law.Downward speedTime after release0vt

The slope gets smaller as acceleration decreases. The curve approaches terminal speed without crossing above it under these unchanged conditions.

Downward acceleration approaches zero

Downward acceleration approaches zeroA schematic graph starts at downward acceleration g and falls towards zero while remaining positive. Downward is the positive direction here. The horizontal time axis is at zero acceleration, and no negative acceleration is shown during this unchanged terminal approach. Time has no numerical scale; the graph does not specify a drag law or a numerical settling time.Downward accelerationTime after release0g

As upward drag grows towards weight, downward resultant force and acceleration approach zero. Acceleration stays downward throughout this approach.

The curves show approach to a limit, not a measured settling time. Zero acceleration at terminal speed means constant downward velocity, not rest.

With downward positive, speed approaches a limiting value and acceleration approaches zero. These qualitative curves show the changing trend without assigning exact elapsed times or a particular drag law.

The speed-time gradient is the downward acceleration during this downward motion. The curve becomes less steep as drag grows and tends to a horizontal line at terminal speed. The body does not stop there. The unchanged-condition model does not predict a speed overshoot followed by an upward acceleration.

The constant-acceleration equations cannot be used across the whole drag-dominated fall: a = g - D/m changes as D changes. Near terminal motion, acceleration is approximately zero even though gravitational field strength remains nonzero.

Follow the energy as well as the forces

Before terminal speed, decreasing gravitational potential energy supplies both increasing kinetic energy and increasing internal energy of the body and air. The kinetic-energy increase is smaller than the gravitational decrease when energy is dissipated.

At terminal speed, kinetic energy is constant, but the body continues descending. Gravitational potential energy continues to decrease, and energy continues transferring to internal stores. A zero resultant is not zero work by each individual force.

Worked terminal-motion account

Continuing transfer at constant speed

In a supplied model, a body's weight is 600 N and its steady downward speed is 50 m/s. Upward drag is therefore also 600 N.

Power of gravity = 600 × 50 = +30000 W = +30 kW
Power of drag = -600 × 50 = -30000 W = -30 kW

The net kinetic-energy change rate is zero. Gravitational energy decreases at 30 kW, while energy transfers to internal stores of the body and air at the same rate under this model. The two forces' work does not vanish merely because their sum is zero.

Check your understanding

A body of weight 600 N falls at a steady terminal speed of 50 m/s in still air. Buoyancy is negligible. Which energy statement is correct?

  1. The kinetic energy increases at 30 kW, because gravity keeps acting downward on the falling object.
  2. No energy transfers because the resultant force is zero.
  3. Gravity supplies 30 kW and drag removes 30 kW to internal stores, so the kinetic energy stays constant.
Show the answer

Gravity supplies 30 kW and drag removes 30 kW to internal stores, so the kinetic energy stays constant.

Gravity power is 600 x 50 = 30000 W. The equal 600 N drag has power -30000 W on the falling body. Gravitational potential energy continues to decrease while internal energy of the body and air increases in this model.

If the drag conditions change

Opening a parachute changes the relationship between drag and speed. Immediately after opening, the upward drag may exceed weight while the body is still moving down. The upward resultant slows its downward motion; it does not make velocity reverse instantly.

As speed decreases, drag falls towards a new balance at a lower terminal speed. With downward positive, the slowing stage has negative acceleration. This changed-condition example differs from the earlier approach to terminal speed for an unchanged body.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A small ball is dropped from rest and falls through air, reaching terminal velocity. Sketch its velocity-time graph and explain its shape. [3]

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