C. Forces · ForcesFree Fall, Air Resistance and Terminal Velocity
C2.6 C2.3 C3.1About 70 minutes
Learning goals- I can explain, using Newton's second law, why all objects in free fall have the same acceleration, g.
- I can solve free-fall problems by combining Fg = mg and the kinematics equations.
- I can explain how air resistance changes the net force and acceleration of a falling object.
- I can describe terminal velocity and the factors that affect it.
Words to know
free fall
Motion in which gravity is the only force acting on an object. Its acceleration is g = 9.8 m/s² down, whether it is moving up, down or at the top of its path.
acceleration due to gravity (g)
The acceleration of a freely falling object: 9.8 m/s² down near Earth's surface. Numerically equal to the gravitational field strength, 9.8 N/kg.
air resistance (drag)
The force of air on an object moving through it, opposite to the motion. It increases with speed and with cross-sectional area.
terminal velocity
The constant, maximum velocity a falling object reaches when air resistance becomes equal to its weight, so Fnet = 0.
A hammer and a feather
In 1971, astronaut David Scott stood on the Moon during the Apollo 15 mission, holding a hammer in one hand and a feather in the other. He let go of both at the same moment, and a television camera recorded them hitting the ground together. On the Moon there is no air.
On Earth a feather drifts down slowly. But that is because of air resistance, not because gravity treats it differently. In this lesson you will see why, using Newton's second law.
Free fall: gravity only
An object is in free fall when gravity is the only force on it. Its FBD has one arrow: Fg = mg, down. By Newton's second law:
Fnet = ma → mg = ma → a = g
The mass cancels. A heavy object has a bigger gravitational force, but it also has more inertia, in exactly the same proportion. So every object in free fall near Earth accelerates at 9.8 m/s² down.
Free fall includes objects thrown upward: a ball rising after you toss it is still in free fall (it is slowing down at 9.8 m/s² each second). At the top, its velocity is zero for an instant, but its acceleration is still 9.8 m/s² down.
To solve free-fall problems, use a = −9.8 m/s² (up as +) or +9.8 m/s² (down as +) in the kinematics equations from the Kinematics unit.
Worked exampleExample 1: A dropped ball
A ball is dropped from rest from a 20 m high balcony. Ignoring air resistance, how long does it take to reach the ground, and how fast is it moving then?
- Down is +. v₁ = 0, a = 9.8 m/s², Δd = 20 m.
- Δd = v₁Δt + ½aΔt² → 20 = 4.9Δt², so Δt² = 4.08, Δt ≈ 2.0 s.
- v₂ = v₁ + aΔt = 0 + 9.8 × 2.02 ≈ 19.8 m/s down (about 71 km/h).
- The ball's mass was not needed.
About 2.0 s; about 20 m/s down.
Worked exampleExample 2: Same acceleration, different forces
A 5.0 kg bowling ball and a 0.50 kg ball are dropped together in a vacuum. Find the gravitational force and acceleration of each.
- Bowling ball: Fg = 5.0 × 9.8 = 49 N. a = 49 ÷ 5.0 = 9.8 m/s².
- Small ball: Fg = 0.50 × 9.8 = 4.9 N. a = 4.9 ÷ 0.50 = 9.8 m/s².
- Ten times the force, but ten times the mass: same acceleration. They land together.
49 N and 4.9 N; both accelerate at 9.8 m/s² down.
Adding air resistance
In real air, a falling object also feels air resistance, pointing up (opposite to its downward motion). Air resistance increases with:
- speed (roughly with the square of the speed at everyday speeds),
- cross-sectional area (how big the object looks from below),
- shape (streamlined shapes have less drag) and air density.
Now Fnet = Fg − Fair (down as +), and a = (mg − Fair) ÷ m, which is less than g.
Follow a skydiver:
- Just after jumping: her speed is small, so Fair ≈ 0. a ≈ 9.8 m/s².
- Speeding up: Fair grows, so Fnet and a get smaller. She still speeds up, but more and more slowly.
- Terminal velocity: Fair has grown until it equals mg. Fnet = 0, so a = 0 and her velocity stays constant (first law). Falling belly-down, this is roughly 55 m/s, about 200 km/h.
- Parachute opens: the area jumps, so Fair becomes larger than mg. The net force points up, and she slows down until a new, much lower terminal velocity of only a few metres per second is reached, slow enough to land safely.
Two objects of the same size and shape but different masses: the heavier one needs more air resistance to balance its weight, so it has a higher terminal velocity and lands first. That is why a crumpled sheet of paper beats a flat one, and why a stack of coffee filters falls faster than a single filter.
Worked exampleExample 3: A skydiver's acceleration
A 75 kg skydiver is falling. (a) Find her acceleration when the air resistance is 400 N. (b) What is the air resistance at her terminal velocity?
- Fg = mg = 75 × 9.8 = 735 N down. Down is +.
- (a) Fnet = 735 − 400 = 335 N. a = 335 ÷ 75 ≈ 4.5 m/s² down. She is still speeding up, but at less than half of g.
- (b) At terminal velocity a = 0, so Fair = Fg = 735 N up.
(a) About 4.5 m/s² down. (b) 735 N.
Velocity of a skydiver falling belly-down (solid curve). At first it rises with slope 9.8 m/s², like free fall (dashed line). Then it bends over and levels off at a terminal velocity of roughly 55 m/s (about 200 km/h).Just for funWhy the falcon feather? On Apollo 15, David Scott chose a falcon feather for his hammer-and-feather drop because the lunar module he had landed in was named Falcon.
Hands-on activityCoffee-filter terminal velocity
You need: 6 to 8 identical paper cone coffee filters, a tape measure, a phone stopwatch (slow-motion video is even better), and a helper.
- Hold one filter, open side up, at a height of 2.0 m and drop it. It reaches terminal velocity almost at once. Time the fall. Repeat 3 times and average.
- Nest 2 filters together (same shape and size, twice the mass) and repeat. Then 4, then 6.
- Calculate the average speed for each: v = 2.0 m ÷ t. Since the filters reach terminal velocity quickly, this is close to the terminal velocity.
- Graph terminal velocity against number of filters. Does doubling the mass double the terminal velocity? (For drag that grows with v², you should find v ∝ √m: 4 filters fall about twice as fast as 1.)
- Explain using FBDs why the heavier stacks fall faster even though, in a vacuum, they would all fall together.
Stand on the floor, not on a chair, to drop them, or have a grown-up hold the filters up high.
Your turn
1What is the only force on an object in free fall?
2Use Newton's second law to show that a 2.0 kg rock and a 6.0 kg rock in free fall have the same acceleration.
a = Fg ÷ m = mg ÷ m = g. For 2.0 kg: 19.6 ÷ 2.0 = 9.8 m/s². For 6.0 kg: 58.8 ÷ 6.0 = 9.8 m/s². Same acceleration.
3A stone is dropped down a well and hits the water after 1.5 s. How deep is the well (to the water)? How fast is the stone going?
Δd = ½(9.8)(1.5)² = 11 m (11.0 m). v = 9.8 × 1.5 = 14.7 m/s (about 15 m/s).
4A ball is thrown straight up. What is its acceleration (a) on the way up, (b) at the top, (c) on the way down? Ignore air.
9.8 m/s² down in all three cases: gravity is the only force throughout.
5A 60 kg skydiver feels 300 N of air resistance. Find her acceleration.
Fg = 588 N. Fnet = 588 − 300 = 288 N down. a = 288 ÷ 60 = 4.8 m/s² down.
6What is true about the forces on an object falling at terminal velocity?
Air resistance equals the weight, so Fnet = 0 and the velocity is constant.
More practice: warm-up
1What is the acceleration of a freely falling object near Earth's surface?
2In which direction does air resistance act on a falling object?
Up (opposite to its motion).
3Does air resistance increase or decrease as a falling object speeds up?
4What is the net force on a skydiver at terminal velocity?
5Why did the hammer and feather land together on the Moon?
There is no air on the Moon, so both were in free fall with the same acceleration.
More practice: on your own
1A ball is thrown straight up at 15 m/s. How long does it take to reach the top, and how high does it go? (Ignore air.)
Up +, a = −9.8. Top: 0 = 15 − 9.8Δt, Δt ≈ 1.5 s (1.53 s). Height: 0 = 15² − 2(9.8)Δd, Δd = 225 ÷ 19.6 ≈ 11 m (11.5 m).
2A 0.060 kg tennis ball falls with 0.30 N of air resistance acting on it. Find its acceleration.
Fg = 0.588 N. Fnet = 0.588 − 0.30 = 0.288 N. a = 0.288 ÷ 0.060 = 4.8 m/s² down.
3Explain why a skydiver slows down when the parachute opens, even though she is still falling.
The open parachute has a huge area, so air resistance becomes greater than her weight. The net force points up, opposite to her downward velocity, so she slows down until the forces balance again at a lower terminal velocity.
4Two balls of the same size are dropped from a tall building: one is solid steel, one is hollow plastic. Which reaches the ground first, and why?
The steel ball. Same size and shape means the same air resistance at the same speed, but the steel ball's weight is much bigger, so air resistance takes away a much smaller fraction of its net force; it has a higher acceleration and a higher terminal velocity.
5Sketch (describe) the v-t graph of a skydiver from the jump until just before the parachute opens.
A curve starting at the origin with slope 9.8 m/s², becoming less steep as speed increases, and levelling off at terminal velocity (a horizontal line).
Challenge
1A 0.50 kg ball dropped from a 45 m tower takes 3.4 s to reach the ground. Find the average acceleration and the average air resistance. (Assume the acceleration is constant.)
Δd = ½aΔt²: a = 2 × 45 ÷ 3.4² = 90 ÷ 11.56 ≈ 7.8 m/s². Fnet = 0.50 × 7.8 = 3.9 N. Fg = 4.9 N. Average Fair ≈ 4.9 − 3.9 = 1.0 N (up).
2At terminal velocity, drag is roughly proportional to v². A 70 kg skydiver has a terminal velocity of 54 m/s. Estimate the terminal velocity of a 105 kg skydiver of the same size and shape.
At terminal velocity mg ∝ v², so v ∝ √m. v = 54 × √(105 ÷ 70) = 54 × √1.5 ≈ 54 × 1.22 ≈ 66 m/s.
3A student says, "At the top of its flight, a ball thrown upward has zero velocity, so it has zero acceleration and no net force." Explain the error.
Velocity and acceleration are different things. At the top, gravity still acts, so Fnet = mg down and a = 9.8 m/s² down. If the acceleration really were zero there, the ball would stay hanging in mid-air (first law)! Instead its velocity is changing from upward to downward at that instant.
Quiz
1. Ignoring air resistance, a 10 kg rock and a 1 kg rock dropped together:
2. A ball thrown straight up is at the top of its path. Its acceleration is:
3. As a falling skydiver speeds up (before terminal velocity), her acceleration:
4. At terminal velocity, the air resistance on a 60 kg skydiver is about:
5. An object dropped from rest falls freely for 3.0 s. Its speed is about:
6. A stack of 4 coffee filters falls faster than 1 filter because:
Big idea: Relating gravitational force to the acceleration of falling objects (C2.6), how net force changes as air resistance grows (C2.3), and the effect of air resistance on velocity (C3.1).
How to help: Do the coffee-filter activity together; it is quick and convincing. Ask your teen to draw the free-body diagram at the start of a fall and at terminal velocity.
Common difficulties:
- Believing heavier objects fall faster in a vacuum.
- Thinking acceleration is zero at the top of a throw.
- Confusing terminal velocity (constant speed) with stopping.
- Thinking the skydiver goes up when the parachute opens. She keeps falling, but slows down.