3.5 The complex behaviour of a falling parachutist

How and why does a parachute work? What does it do?

This is an interesting case because it involves an acceleration, a deceleration and two terminal velocities!

The influence of any side-wind is ignored in this analysis.

These diagrams show the relative magnitude of the two forces acting on a descending parachutist.

Note the relative size and direction of the arrows. The weight force F2 is constant.

1. When the parachutist opens the parachute, there is an immediate large drag force due to big increase in air resistance (air friction).

2. As the parachutist slows down (decelerates) the air resistance (F1) reduces and continues to do so until F1 = F2.

3. When drag force F1 equals the weight force F2, the parachutist is descending at a steady speed - a terminal velocity.

Remember the drag force on an object due to friction (air resistance), increases with speed i.e. as more air brushes over the surface of the object in the same time.

We now look at the whole sequence of events after the parachutist has jumped out of the aircraft

speed/velocity-time graph for a parachute descent

  weight due to gravity W↓ and the drag effect due to air resistance D↑

(I'm ignoring the minor effect of upthrust U↑ that applies to any fluid - gas or liquid)

The force vectors operating in the jump

(1) The parachutist jumps out of the aircraft and immediately accelerates due to gravity (weight force W↓), but the parachute is not opened yet, so the parachutist is in free fall!

Between (1) to (3) acceleration is taking place, but although W↓ is constant, the drag force D↑ due to air friction (air resistance) is increasing, so the acceleration is decreasing (gradient decreasing).

At (3) the terminal velocity is first reached when the drag force D↑ = weight W of the parachutist.

 Its apparently great to do a spot of sky diving first at ~120 mph (~54 m/s) with arms and legs stretched out (increased surface area) and ~200 mph (~89 m/s) if body compact (minimum surface area).

In the graph I've assumed its one or the other!

Under the influence of the Earth's gravitational field this first terminal velocity depends on the shape of your body, but its pretty fast!!!.

Between (3) to (5) the descent continues at the (1st) constant terminal velocity because the resultant force is zero.

At (5) the parachutist opens the parachute to increase air resistance. The wide area brushing through the air produces a massive increase in air resistance (friction), so the drag effect D↑ massively increases, and is far greater than W, so rapid deceleration immediately takes place.

Between (5) to (7) the deceleration continues and as the parachutist decreases in speed, so the drag forces decreases too.

At (7) the drag force once again equals the weight force and the parachutist attains a 2nd, somewhat slower, terminal velocity.

Between (7) and (8) the parachutist descends far more slowly with constant speed to land safely at ~15 mph, again the resultant force is zero.