When an object is thrown straight up or dropped, gravity is the only force acting (ignoring air resistance). This produces uniform acceleration and allows us to use the equations of motion developed in Grade 11. Mastering sign conventions and graph interpretation are the two keys to scoring full marks in this topic.
2.1 Free Fall and Gravitational Acceleration
Definition
Free fall
The motion of a body where the only force acting on it is gravity. An object in free fall accelerates downward at g = 9.8 m·s⁻².
Definition
Gravitational acceleration (g)
The acceleration due to gravity near the Earth's surface is g = 9.8 m·s⁻² directed downward.
In free fall, every object — regardless of mass — accelerates at exactly 9.8 m·s⁻² downward (ignoring air resistance). This is why a feather and a hammer hit the ground simultaneously on the Moon where there is no air. Near Earth's surface air resistance becomes significant for light, large objects, but in CAPS problems we always assume true free fall.
Exam Tip
EXAM TIP: Choose your positive direction ONCE at the start and stick to it. The most common convention is: upward = positive, downward = negative. With this choice, g = −9.8 m·s⁻² (downward acceleration is negative). Write your convention at the top of the solution.
Formula
Gravitational acceleration (sign convention: up = positive)
a = acceleration (m·s⁻²), negative sign indicates downward direction when upward is taken as positive
SI unit: m·s⁻²
Note
At the highest point of a vertical projectile's path: velocity = 0 m·s⁻¹ (object momentarily stationary), but acceleration = 9.8 m·s⁻² downward (gravity never switches off). This is one of the most frequently tested concepts in Grade 12.
Worked Example
A stone is dropped from rest from the top of a cliff 45 m high. Calculate (a) the time it takes to reach the bottom and (b) the velocity just before impact. Take downward as positive.
Given
- v_i = 0 m·s⁻¹ (dropped from rest)
- a = g = +9.8 m·s⁻² (downward = positive)
- s = +45 m (downward)
Find
(a) time t (b) v_f
Solution
- 1(a) s = v_i·t + ½·a·t² → 45 = 0 + ½×9.8×t² → t² = 45/4.9 = 9.18 → t = 3.03 s
- 2(b) v_f² = v_i² + 2as = 0 + 2×9.8×45 = 882 → v_f = 29.7 m·s⁻¹
Practice Question
Explain why a heavy iron ball and a tennis ball dropped simultaneously from the same height both hit the ground at the same time (assume no air resistance).
(3 marks)
2.2 Equations of Motion for Vertical Projectiles
The Three Equations of Motion (vertical free fall)
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
Formula
Equation 1 — velocity-time
v = final velocity (m·s⁻¹), u = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), t = time (s)
SI unit: m·s⁻¹
Formula
Equation 2 — displacement-time
s = displacement (m), u = initial velocity (m·s⁻¹), t = time (s), a = acceleration (m·s⁻²)
SI unit: m
Formula
Equation 3 — velocity-displacement
v = final velocity (m·s⁻¹), u = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), s = displacement (m)
SI unit: m·s⁻¹
Exam Tip
EXAM TIP: Write down all known quantities and the unknown before selecting an equation. Choose the equation that contains exactly your three known quantities and the one unknown. This avoids errors from using the wrong equation.
Worked Example
A ball is thrown vertically upward at 15 m·s⁻¹. Take upward as positive, g = 9.8 m·s⁻². Find: (a) the maximum height reached, (b) the total time in the air before returning to the same launch height.
Given
- u = +15 m·s⁻¹ (upward)
- a = −9.8 m·s⁻² (downward)
- At maximum height: v = 0 m·s⁻¹
Find
(a) max height s_max (b) total time t_total
Solution
- 1(a) v² = u² + 2as → 0 = 15² + 2×(−9.8)×s → s = 225/19.6 = 11.48 m
- 2(b) Time to top: v = u + at → 0 = 15 − 9.8t → t_up = 15/9.8 = 1.53 s
- 3By symmetry (same launch and landing height): t_total = 2×1.53 = 3.06 s
Watch Out
COMMON MISTAKE: Students set a = +9.8 m·s⁻² when they have chosen upward as positive. With upward positive, g = −9.8 m·s⁻². Check your sign for 'a' every time you start a new calculation.
Practice Question
A ball is thrown upward at 20 m·s⁻¹. Taking upward as positive, calculate the velocity of the ball after 3 s.
(3 marks)
Practice Question
A stone is thrown upward at 12 m·s⁻¹ from the edge of a building 20 m tall. Taking upward as positive, calculate the displacement of the stone when it hits the ground at the base of the building.
(6 marks)
2.3 Graphs of Vertical Projectile Motion
Three types of graphs describe vertical projectile motion: position–time (s–t), velocity–time (v–t), and acceleration–time (a–t). Each graph has a distinctive shape that must be recognised and sketched correctly in exams.
Graphs for a Ball Thrown Vertically Upward (up = positive)
| Property | Graph type | Key features |
|---|---|---|
| Position–time (s–t) | Parabola | Upward-opening parabola; peak at maximum height; symmetric if ball returns to start |
| Velocity–time (v–t) | Straight line with negative gradient | Starts at +u; crosses zero at top; ends at −v_f; gradient = −9.8 m·s⁻² |
| Acceleration–time (a–t) | Horizontal line | Constant at −9.8 m·s⁻² throughout entire motion |
Exam Tip
EXAM TIP: On the v–t graph, the area between the graph and the horizontal axis gives displacement (not distance). Areas above the axis are positive (upward displacement); areas below are negative (downward displacement).
Worked Example
A ball is thrown upward at 19.6 m·s⁻¹ and caught at the same height. Sketch the v–t graph and use it to find the total time of flight.
Given
- u = +19.6 m·s⁻¹
- a = −9.8 m·s⁻²
- Returns to same height so net displacement = 0
Find
Total time of flight (from graph)
Solution
- 1Time to reach top: t_up = u/|g| = 19.6/9.8 = 2 s
- 2By symmetry, time to fall back = 2 s
- 3Total time = 4 s
- 4On v–t graph: straight line from v = +19.6 at t=0, crosses zero at t=2, reaches v = −19.6 at t=4
Watch Out
COMMON MISTAKE: On the acceleration–time graph many students draw a curve or change the value at the top of the trajectory. The acceleration is always −9.8 m·s⁻² (constant) — it does not change at the highest point.
Practice Question
A ball is thrown upward at 14.7 m·s⁻¹ and caught at the same point. Using upward as positive, (a) sketch the a–t graph for the entire motion, and (b) state the value of acceleration at the highest point.
(4 marks)
Practice Question
The area under a v–t graph for a vertically thrown ball from t = 0 to t = 2 s is +20 m, and from t = 2 s to t = 4 s is −20 m. What is the displacement of the ball after 4 s? What is the total distance travelled?
(3 marks)