Know that the acceleration of free fall for an object near to the Earth is approximately constant and that this is equivalent to the acceleration of free fall

1.2 Motion – Acceleration of Free Fall

Learning Objective

Understand that the acceleration of free fall for an object near the Earth’s surface is effectively constant and is denoted by g. Use the value of g (≈ 9.8 m s⁻², often rounded to 10 m s⁻²) together with the kinematic equations to solve motion problems in which gravity is the only force acting.

Key Concepts

• Free fall – Motion of a body under the action of gravity alone. Air resistance is ignored unless the question states otherwise (Cambridge requirement).
• Acceleration due to gravity (g)

  • Constant vector directed vertically downwards.
  • Official IGCSE value: g = 9.8 m s⁻².
  • Rounded value for quick work: g = 10 m s⁻² (permitted).

• Why g can be treated as constant

  From Newton’s law of gravitation,

  \[

  g=\frac{GM}{R^{2}}

  \]

  where G is the universal gravitational constant, M the Earth’s mass and R the distance from the Earth’s centre.

  For motions close to the surface, the change in distance is tiny:

  \[

  \frac{\Delta g}{g}\approx 2\frac{h}{R}\qquad(R\approx6.4\times10^{6}\,\text{m})

  \]

  If the height h is a few kilometres (h ≪ R), the change in g is < 1 % and may be ignored. Hence g is taken as constant for all IGCSE questions.

• Sign convention

  • Most IGCSE work takes the downwards direction (the direction of g) as positive.
  • If you choose upwards as positive, write the acceleration as a = –g and keep the sign consistent for all vertical quantities.

• Vector vs scalar

  • Velocity – vector (magnitude + direction). Its scalar magnitude is called speed.
  • Acceleration – vector. In free fall the acceleration vector is always g (downwards), regardless of the object’s instantaneous direction of motion.

• Significant figures – Give the final answer with the same number of significant figures as the data supplied (e.g., 20 m → 2 s to 2 s.f.).

Graphical Interpretation (Syllabus Requirement)

Area under a speed–time graph gives the distance travelled:

• For a straight‑line speed–time graph, distance = (average speed) × time = ½ (g t) · t = ½ g t², which is exactly the equation s = ½gt².

Kinematic Equations for Constant Acceleration (including free fall)

When the acceleration a is constant (here a = g), the four equations below are valid. Choose the one that contains the quantities you know.

1. v = u + at
2. s = ut + \tfrac12 at^{2}
3. v^{2} = u^{2} + 2as
4. s = \dfrac{(u+v)}{2}\,t (average‑velocity form)

Symbols (SI units):

• u – initial velocity (m s⁻¹)
• v – final velocity (m s⁻¹)
• a – acceleration (m s⁻²); for free fall a = g
• t – time (s)
• s – displacement (m); positive downwards if that is the chosen positive direction

Typical Free‑Fall Situations (downwards taken as positive)

Worked Examples

Core‑level example

Problem: A stone is dropped from a height of 20 m. Using g = 10 m s⁻², find the time taken to reach the ground.

Given: u = 0, s = 20 m, a = g = 10 m s⁻².

Using s = ut + \tfrac12 gt^{2}:

\[

20 = 0 \times t + \frac12 (10) t^{2}

\;\Longrightarrow\;

20 = 5t^{2}

\;\Longrightarrow\;

t^{2}=4

\;\Longrightarrow\;

t = 2.0\ \text{s}

\]

Answer: 2.0 s (2 s.f.).

Extended‑level vector example

Problem: A ball is projected from the top of a 25 m high cliff with a speed of 12 m s⁻¹ at 30° above the horizontal. Ignoring air resistance, determine:

1. Time taken to reach the ground.
2. Speed just before impact.

Solution outline:

1. Resolve the initial velocity:

   • uₓ = 12\cos30° = 10.4 m s⁻¹ (horizontal, constant)
   • uᵧ = 12\sin30° = 6.0 m s⁻¹ (upwards → uᵧ = –6.0 m s⁻¹ if downwards is positive)

2. Vertical motion (downwards positive):

   \[

   s = ut + \tfrac12 gt^{2}\;\Longrightarrow\;

   25 = (-6.0)t + 4.9t^{2}

   \]

   Solve \(4.9t^{2}-6.0t-25=0\) → \(t = 3.1\ \text{s}\) (positive root).

3. Horizontal displacement (optional): \(sₓ = uₓ t = 10.4 \times 3.1 \approx 32\ \text{m}\).
4. Final vertical velocity:

   \[

   vᵧ = uᵧ + gt = -6.0 + 9.8(3.1) \approx 24.4\ \text{m s}^{-1}

   \] (downwards)

5. Resultant speed:

   \[

   v = \sqrt{vₓ^{2}+vᵧ^{2}}

   = \sqrt{(10.4)^{2}+(24.4)^{2}}

   \approx 26.5\ \text{m s}^{-1}

   \]

Answers (2 s.f.):

a) 3.1 s**

b) 26 m s⁻¹**

Common Misconceptions (and Clarifications)

• “Free fall only means falling straight down.” – Any motion under gravity alone (upwards, downwards or horizontal) is free fall.
• “Acceleration increases as the object speeds up.” – The acceleration due to gravity remains constant; only the velocity changes.
• “Air resistance can always be ignored.” – In IGCSE questions you may ignore it unless the question explicitly mentions it. When it is mentioned, the net acceleration is reduced.
• “The sign of g changes when the object moves upwards.” – g always points downwards. If you choose upwards as positive, write the acceleration as a = –g.
• “Speed and velocity are the same thing.” – Velocity is a vector (has direction); speed is the scalar magnitude of that vector.

Practice Questions

1. A ball is thrown vertically upwards with an initial speed of 15 m s⁻¹. Using g = 10 m s⁻², calculate:

   1. The time to reach the highest point.
   2. The maximum height above the launch point.

2. An object is dropped from a height of 45 m. Determine the speed just before it hits the ground (use g = 9.8 m s⁻²).
3. A stone is thrown downwards from a cliff with an initial speed of 5 m s⁻¹. If it hits the water 3 s later, find the height of the cliff (use g = 10 m s⁻²).
4. Extended (vector) question: A projectile is launched from ground level with a speed of 20 m s⁻¹ at 40° above the horizontal. Ignoring air resistance, find:

   1. The total time of flight.
   2. The horizontal range.

   (Use g = 9.8 m s⁻² and give answers to 3 significant figures.)

Suggested Diagram

Summary

The acceleration of free fall, g ≈ 9.8 m s⁻², can be regarded as constant for motions where the height is much smaller than the Earth’s radius. Treating g as a constant lets us apply the four kinematic equations, interpret distance‑time and speed‑time graphs, and use vector reasoning to solve a wide range of Core and Extended IGCSE physics problems.