Lesson 1.6 · Unit 1 · HS-PS2-1
Everything falls at the same rate, and that is not obvious
A stone falls faster than a feather, which makes the claim that all objects fall alike look plainly false. It took a careful argument to see past the air, and the argument is worth following because it shows how a physical law can be extracted from observations that seem to contradict it.
- Free fall means gravity is the only force acting. Air resistance excluded, whether the object is moving up, down or momentarily at rest.
- Near Earth's surface the acceleration is \( 9.8\ \text{m}/\text{s}^2 \) downward, for every object regardless of mass.
- The kinematic equations apply directly, with \( a \) replaced by \( \pm g \) according to the sign convention chosen.
- An object thrown upward is in free fall the whole time, including at the top of its flight.
- The motion is symmetric: time up equals time down, and the speed at any height going up equals the speed at that height coming down.
- Mass does not appear in any of the equations, which is why it cannot affect the result.
- Air resistance breaks all of this, and does so more for light or broad objects than for dense compact ones.
Where students lose marks: switching sign conventions partway through a problem, usually by taking down as positive during the fall and up as positive during the rise. Choose once, write it at the top, and let every sign in the problem follow from it.
The problem. (a) Give Galileo's argument that heavy and light objects must fall together. (b) A stone is dropped from 45 m. Find the time to land and the impact speed. (c) A ball is thrown straight up at 19.6 m/s. Find the time to the top and the maximum height. (d) Explain why a feather falls slowly without contradicting any of this.
Step one: set up (a). Suppose, as Aristotle held, that heavier objects fall faster. Take a heavy stone H and a light stone L, so H falls faster than L on its own. Now tie them together and drop them.
Step two: draw out the contradiction. Two answers follow from the same assumption. The lighter stone falls more slowly, so it drags on the heavier one, and the pair should fall slower than H alone. But the pair is heavier than H alone, so by the assumption it should fall faster than H. The same premise gives both "slower than H" and "faster than H", so the premise cannot stand. Galileo put the argument in dialogue in Dialogues Concerning Two New Sciences (Crew and de Salvio translation, 1914), where Salviati leads Simplicio to exactly this result. Note that no experiment was required: the assumption defeats itself.
Step three: set up (b). Take down as positive, so \( a = +9.8\ \text{m}/\text{s}^2 \), \( v_0 = 0 \), \( \Delta x = +45\ \text{m} \). Find the time with the equation that omits final velocity: \( 45 = 0 + \tfrac{1}{2}(9.8)t^2 \), so \( t^2 = \dfrac{90}{9.8} = 9.184 \) and \( t = 3.03\ \text{s} \).
Step four: find the impact speed. \( v = v_0 + at = 0 + 9.8(3.03) = 29.7\ \text{m/s} \). Check independently with the equation that omits time: \( v^2 = 0 + 2(9.8)(45) = 882 \), and \( \sqrt{882} = 29.7\ \text{m/s} \) ✓
Step five: set up (c). Now take up as positive, so \( v_0 = +19.6\ \text{m/s} \) and \( a = -9.8\ \text{m}/\text{s}^2 \). At the highest point the velocity is momentarily zero, which is the condition that locates the top. \( 0 = 19.6 + (-9.8)t \), so \( t = \dfrac{19.6}{9.8} = 2.0\ \text{s} \).
Step six: find the maximum height. \( v^2 = v_0^2 + 2a\Delta x \) with \( v = 0 \): \( 0 = 19.6^2 + 2(-9.8)\Delta x \), so \( \Delta x = \dfrac{384.16}{19.6} = 19.6\ \text{m} \). Note the acceleration is still \( -9.8 \) at the top, as lesson 1.4 argued. Zero velocity is not zero acceleration.
Step seven: use the symmetry as a check. The ball takes 2.0 s to rise 19.6 m. Falling 19.6 m from rest should take \( t = \sqrt{\dfrac{2(19.6)}{9.8}} = \sqrt{4.0} = 2.0\ \text{s} \) ✓ And its speed on return to the throwing height should be 19.6 m/s downward: \( v = 9.8(2.0) = 19.6\ \text{m/s} \) ✓ The symmetry is exact in free fall and is worth using as a check on any answer, but it fails as soon as air resistance enters, because the drag reverses direction with the motion while gravity does not.
Step eight: answer (d). A falling feather is not in free fall, because gravity is not the only force on it. Air resistance is also acting. Why the feather suffers more. Drag depends on surface area and speed, while weight depends on mass. A feather has a great deal of area for very little mass, so the drag force becomes comparable to its weight almost immediately, and it reaches terminal speed within a fraction of a second. A stone has little area for a great deal of mass, so drag stays small compared with its weight for a long time, and it behaves almost exactly as the free-fall equations say. The prediction this makes, and the test. Remove the air and the difference should vanish entirely. In 1971, on the airless surface of the Moon, Apollo 15 commander David Scott dropped a hammer and a falcon feather together and they struck the ground at the same moment, on camera (NASA). The law had been right for three centuries; the air had been hiding it.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \) and ignore air resistance unless told otherwise.
- Give the acceleration of a freely falling object near Earth's surface.
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\( 9.8\ \text{m}/\text{s}^2 \) downward
- Does a heavier object fall faster in a vacuum?
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No
- What is the velocity of a ball at the highest point of its flight?
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Zero
- What is its acceleration at that point?
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\( 9.8\ \text{m}/\text{s}^2 \) downward
- An object is dropped from rest. Find its speed after 3.0 s.
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\( 9.8 \times 3.0 \). 29.4 m/s
- A ball is dropped from 20 m. Find the time to land and the impact speed.
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Down positive, \( v_0 = 0 \). \( 20 = \tfrac{1}{2}(9.8)t^2 \), so \( t^2 = \dfrac{40}{9.8} = 4.082 \) and \( t = 2.02\ \text{s} \). \( v = \sqrt{2(9.8)(20)} = \sqrt{392} = 19.8\ \text{m/s} \). Check: \( v = 9.8(2.02) = 19.8\ \text{m/s} \) ✓ 2.02 s and 19.8 m/s
- A ball is thrown up at 24.5 m/s. Find the maximum height and the total time in the air.
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Up positive, \( a = -9.8\ \text{m}/\text{s}^2 \). Height: \( \Delta x = \dfrac{24.5^2}{2(9.8)} = \dfrac{600.25}{19.6} = 30.6\ \text{m} \). Time up: \( \dfrac{24.5}{9.8} = 2.5\ \text{s} \). Total time, by symmetry: \( 2 \times 2.5 = 5.0\ \text{s} \). The symmetry halves the work, and it is exact here because air resistance is being ignored. 30.6 m and 5.0 s
- Two balls are released from the same height, one dropped and one thrown horizontally.
Which lands first?
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They land together. The horizontal throw gives the second ball a horizontal velocity, and horizontal velocity has no effect on vertical motion. Both start with zero vertical velocity and both have \( 9.8\ \text{m}/\text{s}^2 \) downward, so both satisfy the same vertical equation. This is the independence of components that lesson 1.7 develops, and it is one of the least intuitive results in the unit. They land at the same time
- Explain why the mass of a falling object does not affect its acceleration, given that a
heavier object is pulled harder by gravity.
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Because the greater pull is exactly offset by the greater resistance to being accelerated, and the two effects cancel completely. Set it out with the second law, which unit 2 states formally as \( F = ma \). The gravitational force on a mass \( m \) is its weight, \( F = mg \). The acceleration that force produces is \( a = \dfrac{F}{m} = \dfrac{mg}{m} = g \). The mass cancels. Doubling the mass doubles the force, and doubling the force on a doubled mass gives the same acceleration. Why the cancellation is not trivial. The \( m \) in \( mg \) measures how strongly gravity pulls on the object. The \( m \) in \( F = ma \) measures how strongly the object resists being accelerated by any force at all. There is no obvious reason those two quantities should be the same number, and experiment says they are, to extraordinary precision. That equality has a name and consequences. It is the equivalence of gravitational and inertial mass, and its exactness was the observation Einstein built general relativity on. A student meeting it here is meeting the starting point of a much later theory. What would happen if they differed. Objects made of different materials would fall at measurably different rates, and every free-fall equation in this unit would need a material-dependent correction. Experiments looking for exactly that have found nothing to the limits of their precision. The extra pull is exactly canceled by the extra inertia
- A ball is thrown straight up from the ground at 19.6 m/s from a 30 m building's roof.
Find when it lands at the base of the building and its speed on landing.
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Set the frame carefully. Take up as positive with the origin at the roof. Then: \( v_0 = +19.6\ \text{m/s} \), \( a = -9.8\ \text{m}/\text{s}^2 \), and the ground is at \( \Delta x = -30\ \text{m} \), because it is below the origin. Use the equation with displacement and time. \( -30 = 19.6t + \tfrac{1}{2}(-9.8)t^2 \) \( -30 = 19.6t - 4.9t^2 \) Rearranged into standard form: \( 4.9t^2 - 19.6t - 30 = 0 \). Solve the quadratic. Discriminant: \( (-19.6)^2 - 4(4.9)(-30) = 384.16 + 588 = 972.16 \). \( \sqrt{972.16} = 31.18 \). \( t = \dfrac{19.6 \pm 31.18}{9.8} \). The two roots are \( t = \dfrac{50.78}{9.8} = 5.18\ \text{s} \) and \( t = \dfrac{-11.58}{9.8} = -1.18\ \text{s} \). Reject the negative root. It describes where the ball would have been before it was thrown, which the model does not cover. So the ball lands at 5.18 s. Find the landing speed. \( v = v_0 + at = 19.6 + (-9.8)(5.18) = 19.6 - 50.8 = -31.2\ \text{m/s} \). The negative sign means downward, so the speed is 31.2 m/s. Check with the equation that omits time. \( v^2 = 19.6^2 + 2(-9.8)(-30) = 384.16 + 588 = 972.16 \), so \( v = \sqrt{972.16} = 31.2\ \text{m/s} \) ✓ Notice that check is the discriminant. The same number appeared in the quadratic, which is not a coincidence: both routes solve the same physical relation. Sanity check the size. The ball goes up 19.6 m, comes back to the roof at 19.6 m/s downward after 4.0 s, then falls 30 m more. Falling 30 m from 19.6 m/s takes about 1.2 s and adds speed, so a total near 5.2 s and a landing speed above 19.6 m/s are both what should be expected ✓ A note on the sign of \( \Delta x \). Writing \( +30 \) for the building height is the commonest error here. The ball finishes below its starting point, and with up positive that displacement is negative. Getting this wrong gives a quadratic with no real roots, which at least announces the mistake. 5.18 s, landing at 31.2 m/s