Falling with air resistance: the skydiver's story

By the end of this lesson

By the end you can describe and explain, step by step, how the velocity and acceleration of a falling object change when air resistance acts, and why it ends up at a terminal velocity.

The air is your friend

A skydiver jumps from $3000\ \text{m}$ . With no air, she'd hit about $240\ \text{m s}^{-1}$ by the ground: airliner cruising speed.

With air, she never goes faster than about $55\ \text{m s}^{-1}$ (roughly $200\ \text{km h}^{-1}$ ), however long she falls. Two forces act on her:

her weight $W = mg$ , which is constant;
air resistance $D$ , which increases as her speed increases.

Everything that happens comes from the fight between them.

The story in three acts

	Speed	Air resistance $D$	Resultant force	Acceleration
1. Just released	zero	zero	$W$	$g$
2. Speeding up	increasing	increasing	$W - D$ , decreasing	less than $g$ , decreasing
3. Terminal velocity	constant	$= W$	zero	zero

The crucial bit is act 2. She is still speeding up, just less and less quickly, because each extra bit of speed brings extra drag, which eats into the resultant force.

The one equation you need

Newton's second law, with downwards positive:

$a = \frac{W - D}{m} = g - \frac{D}{m}$

At release, $D = 0$ , so $a = g$ .
As $D$ grows, $a$ falls.
When $D = W$ , $a = 0$ .

Worked example: Halfway through the story

A skydiver of mass $75\ \text{kg}$ is falling. At one instant the air resistance on her is $450\ \text{N}$ . Find her acceleration at that instant, and the air resistance when she reaches terminal velocity. Use $g = 9.81\ \text{m s}^{-2}$ .

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.