Basis Lab

Answers · The physics underneath

Why does an object speed up as it falls?

In shortBecause a falling object has a fixed total of height energy plus motion energy, so every bit of height it loses turns into speed. Gravity converts rather than adds.

There is a fixed total, and only two places for it to sit: height and speed. What the fall takes out of the first it must put into the second, so the object speeds up not because gravity keeps feeding it something new but because its height is being spent. For a mass mm at height hh moving at speed vv, the total is 12mv2+mgh\tfrac12 m v^2 + mgh, and that sum is the same at every instant of the fall.

12mv2+mgh=E=constant\tfrac{1}{2} m v^2 + mgh = E = \text{constant}

Numbers make the trade concrete. Take a 1 kg ball released from rest 5 m above the ground, with g=9.8g = 9.8 m/s² and heights measured from the ground. At release v=0v = 0, so the whole total sits in the height term: E=mgh=1×9.8×5=49E = mgh = 1 \times 9.8 \times 5 = 49 J. Nothing can raise that 49 J or lower it during the fall. It can only be redistributed.

FIG 1
height h (m)mgh (J)½mv² (J)v (m/s)
5.049.000.000.00
3.231.3617.645.94
2.524.5024.507.00
1.817.6431.367.92
0.00.0049.009.90

Read the first two rows as one step. The ball drops 1.8 m, so the height term loses 1×9.8×1.8=17.641 \times 9.8 \times 1.8 = 17.64 J. The speed term gains exactly that, going from 0 to 17.64 J, and solving 12(1)v2=17.64\tfrac12 (1) v^2 = 17.64 gives v=35.28=5.94v = \sqrt{35.28} = 5.94 m/s. The next 1.8 m of fall moves the same 17.64 J across again: height is down to 17.64 J, speed is up to 31.36 J, and v=62.72=7.92v = \sqrt{62.72} = 7.92 m/s. At every height the two columns add to 49.

The speed does not grow evenly with distance, because the moving part depends on the square of the speed. Solving the total for vv gives v=2g (h0−h)v = \sqrt{2g\,(h_0 - h)} with h0=5h_0 = 5 m, so the speed goes as the square root of the distance dropped. Half the drop buys 1/21/\sqrt2 of the final speed: 2.5 m of fall gives v=2×9.8×2.5=49=7.00v = \sqrt{2 \times 9.8 \times 2.5} = \sqrt{49} = 7.00 m/s, against 98=9.90\sqrt{98} = 9.90 m/s at the ground. Most of the speed is picked up early, and the last metre of fall adds less than the first.

The guess that fails

The natural picture is of gravity as a pump: a force that keeps pushing, so the object gets faster each second and ends the fall with more energy than it started with. That fails on the total, which is 49 J at release, 49 J at 3.2 m, and 49 J the instant before impact. Gravity adds nothing. It is the agency by which one term is converted into the other, and the amount convertible was fixed the moment the ball was let go. The same error shows up as expecting the heavy ball to gain speed faster. The mass appears in both terms, so it cancels: mgh=12mv2mgh = \tfrac12 m v^2 gives v=2gΔhv = \sqrt{2g\Delta h} with no mm in it, and a 10 kg ball released beside the 1 kg one passes every height at the same speed, 5.94 m/s at 3.2 m, 9.90 m/s at the ground.

The trade also runs backwards, and its endpoint is where the arithmetic collapses. Throw the ball up at 9.90 m/s and it climbs to 5 m; at the top the speed term is zero and the height term holds the entire 49 J, which is what makes the height easy to compute without ever asking how long the climb takes.

Where the trade stops being exact

With air resistance the sum is no longer fixed, and the height term pays two bills instead of one: the moving part and the drag. The object still speeds up, but the total falls throughout, and once drag balances weight the height term keeps decreasing while the speed term stays level. That is terminal speed, and it is a steady state of a losing system, not a conserved one. Very fast motion breaks the formula earlier: 12mv2\tfrac12 m v^2 stops being the right moving term as the speed approaches that of light.

One more thing is a choice rather than a fact. Where h=0h = 0 sits was picked, and picking the release point instead makes the height term −mgx-mgx for a drop xx, negative throughout, with the speed term cancelling it to a total of zero. The 49 and the 17.64 and the 24.5 are bookkeeping; only their differences, and the fact that they add to a constant, describe the ball. Put the zero on the floor and every row of the table shifts by the same amount while the speeds stay exactly as they were.

Worth keeping

  • In free fall the sum of the motion term and the height term is fixed, so losing height and gaining speed are one event seen from two sides.
  • The speed gained depends on the distance dropped and not on the mass, because the mass appears in both terms and cancels.
  • Air resistance does not change the shape of the trade, it makes the total fall, so the moving part can level off while the height part keeps dropping.

Where this goes next

This question is one step of Motion and What Is Conserved, a course that works through its subject from the beginning. The page that teaches this part of it is Two Kinds of Energy, One Total, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that pageThe whole course

Other questions people ask about this

  1. Why is momentum conserved in a collision when kinetic energy is not?Momentum is conserved because the internal pushes in a collision cancel in pairs, while kinetic energy can be turned into heat and deformation.
  2. Why does a planet speed up when it gets closer to the Sun?A planet speeds up near the Sun because its fixed angular momentum forces its sideways speed up as its distance falls, and gravity does work as it falls inward.
  3. Why is the fastest way down a hill not a straight line?Because time is distance divided by speed, and a steeper path gives the bead more speed early even though it is longer.
  4. Why does a spinning book suddenly start tumbling instead of staying on its axis?Spinning a book about its middle-sized axis is unstable: the conserved quantities still allow it to turn over, and any tiny wobble grows until it flips.
  5. How can light travel through empty space if there's nothing to wave?Light is a self-sustaining ripple of electric and magnetic fields, each one creating the other, so it needs no material to travel through.
  6. How can you predict the shape of a plucked string a moment after you let go?Write the starting corner as a sum of the string's allowed standing shapes, let each shape oscillate at its own rate, then add them back together.

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