Lesson 1.5 · Unit 1 · CED topic 1.3 (gravity near Earth)
Free fall
Drop a coin and a pen from the same height at the same moment and, when air matters little, they hit the floor together. This was the observation that undermined centuries of belief that heavier things fall faster, and it remains one of the clearest examples of constant acceleration in all of physics. Free fall is motion under the influence of gravity alone. It includes objects that are dropped, thrown downward and thrown upward, and it is the simplest place to practice the equations of lesson 1.4, because the acceleration is known before you start.
Describe the position, velocity, and acceleration of an object using representations of that object’s motion.
- Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to \(g = 10\ \text{m/s}^2\).
- For all situations in which a numerical quantity is required for \(g\), the value \(g = 10\ \text{m/s}^2\) will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of \(g = 9.8\ \text{m/s}^2\) or \(g = 9.81\ \text{m/s}^2\).
One acceleration for every falling object
An object in free fall accelerates downward at \(g\), about \(10\ \text{m/s}^2\) near Earth’s surface. Every second that passes, the downward velocity increases by 10 m/s. The value does not depend on the mass, the size or the material of the object, and it does not depend on whether the object is moving up, moving down or momentarily at rest. In this course we use \(g = 10\ \text{m/s}^2\), as the exam does; computing with 9.8 would change the answers by about 2 percent and would also be accepted.
Why the independence of mass? A heavier object feels a larger gravitational force, but it also has more inertia, and the two effects cancel exactly. Unit 2 explains this with Newton’s second law. In everyday life, air resistance makes a feather fall slowly; in a vacuum chamber, and on the Moon when an astronaut dropped a hammer and a feather together in 1971, they land at the same instant. Unless a problem says otherwise, neglect air resistance. That is the “free” in free fall.
Up, down and the sign of \(a\)
Choose a positive direction first. If up is positive, the acceleration is \(a_y = -g = -10\ \text{m/s}^2\) at every instant, including while the object rises. If down is positive, \(a_y = +10\ \text{m/s}^2\) throughout. Both are correct; mixing them is not. Most problems with an object tossed upward are easiest with up positive, so that the starting velocity is positive and the acceleration negative.
Figure 1.5a shows a ball thrown straight up at 20 m/s. On the way up the velocity (an upward arrow) gets shorter by 10 m/s each second, until at the top it is zero. Then it points downward and grows again. The acceleration, drawn as a downward arrow at every position, never changes. This is the picture that corrects the most common error in Unit 1: at the top the velocity is zero but the acceleration is still \(-10\ \text{m/s}^2\). If the acceleration were zero there, the ball would stay at the top forever.
Symmetry of the toss
When an object is thrown up and returns to its starting height, three symmetries follow from the constant acceleration and make many questions fast. The time going up equals the time coming down. The object passes any given height twice, once up and once down, with the same speed in the two directions. And it returns to its launch height with the same speed it had at launch, but now pointing downward. At the top, \(v_y = 0\), so \(t_{\text{up}} = v_0/g\) and the height gained is \(h = v_0^2/(2g)\).
| time (s) | distance fallen from rest (m) | velocity (m/s, down positive) | fell during this second (m) |
|---|---|---|---|
| 1 | 5 | 10 | 5 |
| 2 | 20 | 20 | 15 |
| 3 | 45 | 30 | 25 |
| 4 | 80 | 40 | 35 |
The table shows a dropped object. The distance fallen grows as \(t^2\) (\(\tfrac12 g t^2 = 5t^2\)), while the velocity grows in proportion to \(t\). In successive seconds the object falls 5 m, 15 m, 25 m, 35 m: a steadily growing share of the total, because it is moving faster each second.
A ball is thrown straight up at 20 m/s from the ground. Find the maximum height, the time to reach it, the total time in the air, and the velocity just before it lands.
Axis. Up is positive, so \(v_{y0} = +20\ \text{m/s}\) and \(a_y = -10\ \text{m/s}^2\).
Time to the top. At the top \(v_y = 0\): \(0 = 20 - 10t\), so \(t = 2.0\ \text{s}\).
Maximum height. \(0 = 20^2 + 2(-10)h\), so \(h = 400/20 = 20\ \text{m}\).
Total time. The trip down mirrors the trip up, so the total is \(2 \times 2.0 = 4.0\ \text{s}\). Check with \(y = v_{y0}t + \tfrac12 a_y t^2 = 20(4) - 5(16) = 0\). ✓
Landing velocity. \(v_y = 20 - 10(4.0) = -20\ \text{m/s}\): the same speed as at launch, now directed downward.
A construction worker drops a bolt from the top of a building 45 m high. How long does it take to hit the ground and how fast is it going?
Axis. Down is positive, so \(v_{y0} = 0\), \(a_y = +10\ \text{m/s}^2\), \(\Delta y = +45\ \text{m}\).
Time. \(\Delta y = \tfrac12 a_y t^2\) gives \(t = \sqrt{\dfrac{2(45)}{10}} = \sqrt{9.0} = 3.0\ \text{s}\).
Speed. \(v_y = a_y t = (10)(3.0) = 30\ \text{m/s}\) downward. Check with \(v_y^2 = 2a_y\Delta y = 2(10)(45) = 900\), so \(v_y = 30\ \text{m/s}\). ✓
A student stands at the edge of a 15 m cliff and throws a stone straight up at 10 m/s. The stone misses the cliff on the way down and lands at the bottom. Find the maximum height above the cliff edge, the time of flight and the landing velocity.
Axis. Up is positive, origin at the edge: \(v_{y0} = +10\ \text{m/s}\), \(a_y = -10\ \text{m/s}^2\), landing at \(y = -15\ \text{m}\).
Maximum height. \(0 = 10^2 + 2(-10)h\), so \(h = 5.0\ \text{m}\) above the edge, reached at \(t = 1.0\ \text{s}\).
Time of flight. \(-15 = 10t - 5t^2\), so \(5t^2 - 10t - 15 = 0\) and \(t^2 - 2t - 3 = 0\), giving \((t - 3)(t + 1) = 0\). The physical root is \(t = 3.0\ \text{s}\); the root \(t = -1\ \text{s}\) is a time before the throw and is rejected.
Landing velocity. \(v_y = 10 - 10(3.0) = -20\ \text{m/s}\). The stone passes the edge on the way down at \(-10\ \text{m/s}\) (the symmetry rule) and then gains another 10 m/s over the last 1.0 s.
- Constant \(-g\), always. If a graph, diagram or sentence suggests the acceleration is zero at the top, that is the trap.
- Choose and state the axis. The acceleration of an object in free fall has the sign opposite to the one you assigned to “up.”
- Two roots. A quadratic in \(t\) can give a negative time. Say why you rejected it.
- Neglect air resistance unless told otherwise, and say so when the question asks about the assumptions of your model.
- Mass does not appear. If a question gives masses in a free-fall setting, the masses are distractors; the motion is the same for all of them.
- A rock is dropped from a cliff 80 m high. How long does it fall and how fast is it moving when it lands?
Show answer
\(t = \sqrt{2(80)/10} = 4.0\ \text{s}\). \(v = gt = 40\ \text{m/s}\) downward.
- A ball is thrown straight up at 30 m/s. Find the maximum height and the total time in the air (landing at the same height).
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\(h = 30^2/(2 \cdot 10) = 45\ \text{m}\). Time up \(= 30/10 = 3.0\ \text{s}\); total \(= 6.0\ \text{s}\).
- A stone dropped into a well hits the water 2.0 s later. Ignoring the time sound takes to return, how deep is the well?
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\(\Delta y = \tfrac12 g t^2 = \tfrac12(10)(2.0)^2 = 20\ \text{m}\).
- A ball is thrown straight down at 10 m/s from a window 15 m above the ground. Find the time to land and the landing speed.
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With down positive: \(15 = 10t + 5t^2\), so \(t^2 + 2t - 3 = 0\) and \(t = 1.0\ \text{s}\). \(v = 10 + 10(1.0) = 20\ \text{m/s}\).
- A 1 kg ball and a 5 kg ball are dropped from the same height at the same time. Which reaches the ground first, and why?
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They land together. Near Earth’s surface every freely falling object has the same acceleration, \(g\), independent of mass; the larger gravitational force on the heavier ball is exactly offset by its larger inertia. Air resistance is neglected.
- A ball is thrown upward. Describe its velocity and acceleration at the top of its path, and sketch (in words) the shape of its \(v\)–\(t\) graph.
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At the top the velocity is zero for an instant and the acceleration is \(10\ \text{m/s}^2\) downward. The \(v\)–\(t\) graph, with up positive, is a single straight line with a negative slope of \(-10\ \text{m/s}^2\), starting at the positive launch velocity, crossing the time axis at the top and continuing below the axis as the ball falls.