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ESO Y4B · Physics and Chemistry
Interaction · Kinematics

Free fall

On the Moon, where there is no air, a hammer and a feather dropped at the same time reach the ground together. On Earth they fall differently because of air resistance, but the cause of the motion —gravity— is the same for both. Free fall is, mathematically, a UARM that takes place along the vertical, with an acceleration that does not depend on the mass of the object.

Before you start, make sure you know: Uniformly accelerated rectilinear motion (3rd ESO)
Before you start, make sure you know: Quadratic equations (3rd ESO)

Gravity as a constant acceleration

Animation of a sphere falling from rest towards the ground with constant acceleration: it falls faster and faster and the marks left at equal time intervals get further and further apart. The arrow for g stays fixed on the left. g ≈ 9,8 m/s² ground
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On the Moon, where there is no air, an astronaut dropped a hammer and a feather at the same time and both reached the ground at the same time. The scene is shown in the…

On the Moon, where there is no air, an astronaut dropped a hammer and a feather at the same time and both reached the ground at the same time. The scene is shown in the figure.

hammer feather
  1. a) What does this fact show about falling objects when there is no air resistance?
  2. b) What is the acceleration called with which objects fall on Earth, and what is its approximate value?
Solve
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In free fall, the speed of an object increases by $9,8\ \text{m/s}$ every second. An object is dropped from rest. a) Calculate its speed at $1\ \text{s}$, $2\ \text{s}$…

In free fall, the speed of an object increases by $9,8\ \text{m/s}$ every second. An object is dropped from rest.

  1. a) Calculate its speed at $1\ \text{s}$, $2\ \text{s}$ and $3\ \text{s}$.
  2. b) What acceleration does the object have during the fall, and which letter is used to represent it?
Solve

Speed and distance in a fall

When you measure the fall of a ball from rest, you observe that in $1\ \text{s}$ it covers about $5\ \text{m}$, in $2\ \text{s}$ it has already covered $20\ \text{m}$ and in $3\ \text{s}$ it reaches $45\ \text{m}$: the distance does not increase at the same rate, but shoots up with each second. What mathematical relationship links time with the distance covered?

Look at the $v(t)$ graph: it is a straight line that always rises with the same gradient, $g$. The $x(t)$ graph curves upwards faster and faster, because the velocity keeps increasing.

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A stone is dropped from rest. With $g = 9,8\ \text{m/s}^2$, calculate its velocity after $2\ \text{s}$ of falling using $v = v_0 + g \cdot t$.

A stone is dropped from rest. With $g = 9,8\ \text{m/s}^2$, calculate its velocity after $2\ \text{s}$ of falling using $v = v_0 + g \cdot t$.

Solve
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An object is dropped from rest. With $g = 9,8\ \text{m/s}^2$, calculate the distance it has fallen after $3\ \text{s}$ using $h = \frac{1}{2} g \cdot t^2$.

An object is dropped from rest. With $g = 9,8\ \text{m/s}^2$, calculate the distance it has fallen after $3\ \text{s}$ using $h = \frac{1}{2} g \cdot t^2$.

Solve
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A coin falls from a height of $1,25\ \text{m}$. Take $g \approx 10\ \text{m/s}^2$. a) How long does it take to reach the ground? b) With what speed does it arrive?

A coin falls from a height of $1,25\ \text{m}$. Take $g \approx 10\ \text{m/s}^2$.

  1. a) How long does it take to reach the ground?
  2. b) With what speed does it arrive?
Solve

Vertical projection upwards

Animation of vertical projection: the ball rises while slowing down, stops at the top and falls again while speeding up. The velocity arrow shrinks to zero at the top and then grows downwards, while the gravity arrow does not change. v₀ g = −a (slows down)
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A ball is thrown vertically upwards at $15\ \text{m/s}$. Take $g \approx 10\ \text{m/s}^2$. a) How long does it take to reach its maximum height? Hint: at the highest…

A ball is thrown vertically upwards at $15\ \text{m/s}$. Take $g \approx 10\ \text{m/s}^2$.

  1. a) How long does it take to reach its maximum height? Hint: at the highest point, the velocity is zero.
  2. b) What maximum height does it reach above the release point?
Solve
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A stone is thrown vertically upwards from a bridge at $10\ \text{m/s}$. Take $g \approx 10\ \text{m/s}^2$. a) What maximum height does it reach above the level of the…

A stone is thrown vertically upwards from a bridge at $10\ \text{m/s}$. Take $g \approx 10\ \text{m/s}^2$.

  1. a) What maximum height does it reach above the level of the bridge?
  2. b) How long does it take to pass the level of the bridge again?
Solve

Symmetry of rising and falling

If you throw a coin upwards and time its flight, you will find that it takes almost exactly the same time to rise until it stops as to return to your hand. Why do the rise and the fall look like two identical halves of the same film played forwards and backwards?

Velocity-time graph of a vertical projection: a single descending straight line going from v₀ to −v₀ and crossing the axis at the highest point, with equal times of rise and fall v t v₀ −v₀ t₁ 2t₁ rises (v > 0) falls (v < 0) t rise = t fall
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You throw a ball vertically upwards at $20\ \text{m/s}$ from the ground, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$. a) What maximum height does it reach?…

You throw a ball vertically upwards at $20\ \text{m/s}$ from the ground, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$.

v₀ = 20 m/s h = ?
  1. a) What maximum height does it reach?
  2. b) How much time passes until it returns to your hands?
  3. c) With what velocity does it return? Compare it with the launch velocity and comment on the result.
Solve
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A ball is thrown vertically upwards at $25\ \text{m/s}$ from the ground, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$. a) Calculate the maximum height and…

A ball is thrown vertically upwards at $25\ \text{m/s}$ from the ground, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$.

v₀ = 25 m/s h = 20 m t₁ = ? t₂ = ? h max = ?
  1. a) Calculate the maximum height and the time at which it reaches it.
  2. b) Set up the equation to find at what times it passes through the height of $20\ \text{m}$ and solve it: you will get two solutions.
  3. c) Calculate the velocity of the ball for each of these two solutions.
  4. d) Explain the physical meaning of the two solutions and the relationship between the two velocities.
Solve

Combining free fall and vertical projection: two objects

Two bodies on the same vertical line: one falls from the top and the other is thrown upwards from the ground; the question is when and at what height they meet When and at what height do they meet? x (height) 0 ground (origin) falls starts from rest rises has v₀ upwards
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From the edge of a roof $40\ \text{m}$ above the ground, a ball is thrown vertically upwards at $10\ \text{m/s}$, as shown in the figure. Take…

From the edge of a roof $40\ \text{m}$ above the ground, a ball is thrown vertically upwards at $10\ \text{m/s}$, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$.

v₀ = 10 m/s h = ? 40 m
  1. a) What maximum height does it reach above the roof? And above the ground?
  2. b) How long does it take to go up?
  3. c) Set up the position equation taking the ground as the origin and calculate the total time of flight until the ball reaches the ground.
  4. d) With what velocity does it hit the ground?
Solve
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From the top of a tower $80\ \text{m}$ high, a stone is dropped, and at the same instant, another stone is thrown vertically upwards from the ground at $40\ \text{m/s}$,…

From the top of a tower $80\ \text{m}$ high, a stone is dropped, and at the same instant, another stone is thrown vertically upwards from the ground at $40\ \text{m/s}$, as shown in the figure. Take $g \approx 10\ \text{m/s}^2$.

stone 1 stone 2 v₀ = 40 m/s h = ? 80 m
  1. a) Write the height equations for each stone, $h_1$ and $h_2$, measured from the ground.
  2. b) Calculate the instant at which they cross paths.
  3. c) At what height do they cross paths?
  4. d) Calculate the velocity of each stone at that instant and state whether each one is moving up or down.
Solve
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A hot air balloon ascends at a constant velocity of $5\ \text{m/s}$. When it is $30\ \text{m}$ above the ground, a ballast bag comes loose. Take…

A hot air balloon ascends at a constant velocity of $5\ \text{m/s}$. When it is $30\ \text{m}$ above the ground, a ballast bag comes loose. Take $g \approx 10\ \text{m/s}^2$.

  1. a) Explain why the bag, upon coming loose, continues to move upwards for a moment instead of falling immediately.
  2. b) Taking the ground as the origin and upwards as positive, set up the position equation of the bag and calculate the time it takes to reach the ground.
  3. c) With what velocity does it hit the ground?
Solve

Summary

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