Physics · Lesson 4

Projectile motion, circular motion & SHM

JAMBWAECNECO

What the syllabus asks

JAMB: Motion: projectiles, circular motion, simple harmonic motion

WAEC: Motion: projectile motion (time of flight, maximum height, range), circular motion (angular speed, centripetal acceleration and force), simple harmonic motion (period, frequency, amplitude, energy, simple pendulum, mass–spring)

NECO: Motion: projectile motion (time of flight, maximum height, range), circular motion (angular speed, centripetal acceleration and force), simple harmonic motion (period, frequency, amplitude, energy, simple pendulum, mass–spring)

What you will learn

Analyse horizontal and oblique projectiles by resolving velocity.
Calculate time of flight, maximum height and range.
Relate period, frequency, angular speed and linear speed in circular motion.
Calculate centripetal acceleration and force, and identify what provides it.
Define SHM and use a = −ω²x, v = ω√(A² − x²).
Use the pendulum and spring formulae and describe energy changes in SHM.

Key points

Projectiles: horizontal velocity constant; vertical acceleration g downward; treat separately.
Horizontal launch from height h: t = √(2h/g), range = ut.
Angled launch: T = 2u sinθ/g; H = u²sin²θ/2g; R = u²sin2θ/g; R_max = u²/g at 45°; complementary angles give equal range.
Circular motion: ω = 2π/T = 2πf; v = ωr; a = v²/r = ω²r towards the centre.
Centripetal force mv²/r is supplied by a real force (tension, friction, gravity, reaction).
SHM: a = −ω²x; v = ω√(A² − x²); v_max = ωA; a_max = ω²A; E = ½kA².
Pendulum T = 2π√(l/g) (independent of mass and small amplitude); spring T = 2π√(m/k).
Resonance: large amplitude when driving frequency equals natural frequency.

Start here: What makes a projectile

Kick a football, throw a javelin, or let water spray from a hose, and the object follows a curved path. Any object moving freely through the air under gravity alone is called a projectile, and its path is its trajectory.

The key idea, first understood clearly by Galileo, is that the horizontal and vertical motions are independent. Gravity acts only downward, so it changes only the vertical velocity. If we ignore air resistance, nothing acts horizontally, so the horizontal velocity stays constant.

Horizontal: velocity constant, acceleration 0: x = uₓt
Vertical: acceleration g downward (take g = 10 m/s² unless told 9.8): use v = u + at etc.

A famous demonstration: drop one coin from a table and, at the same instant, flick another coin horizontally off the edge. Both hit the floor at the same time. The horizontal push does not change how fast the coin falls.

Worked example 4.1

A ball is thrown horizontally at 15 m/s from the top of a cliff 20 m high. Find (a) the time to reach the ground, (b) how far from the foot of the cliff it lands. (g = 10 m/s²)

Vertical motion starts with zero vertical velocity: h = ½gt², so 20 = 5t², t = 2 s. We use vertical motion because the time depends only on the fall.
Horizontal: x = 15 × 2 = 30 m.

2 s; 30 m

Worked example 4.2

In example 10.1, find the velocity on landing.

Horizontal component stays 15 m/s.
Vertical: v = 0 + 10 × 2 = 20 m/s downward.
Speed = √(15² + 20²) = 25 m/s; angle below horizontal = tan⁻¹(20/15) = 53.1°.

25 m/s at 53.1° below the horizontal

⚠️ Common mistake: thinking a faster horizontal throw stays in the air longer. The time of fall depends only on the height.

Try it yourself. A plane flying horizontally at 100 m/s at a height of 500 m drops a food parcel. How far ahead of the target must it be released? (g = 10)

Show solution

500 = 5t² → t = 10 s; distance = 1,000 m.

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Practice questions

1. A stone is thrown horizontally at 8 m/s from a height of 45 m. How far does it travel horizontally? (g = 10)

Show worked answer

t = 3 s; x = 24 m.

2. Find the range of a projectile launched at 50 m/s at 45°. (g = 10)

Show worked answer

250 m.

3. A ball is kicked at 25 m/s at 37° (sin 37° = 0.6). Find its time of flight. (g = 10)

Show worked answer

2 × 15/10 = 3 s.

4. A wheel of radius 0.3 m turns at 2 rev/s. Find the linear speed of its rim.

Show worked answer

v = 2π × 2 × 0.3 = 3.77 m/s.

5. Find the centripetal force on a 2 kg mass moving at 6 m/s in a circle of radius 3 m.

Show worked answer

24 N.

Key words

Projectile: An object moving freely under gravity after being launched.
Trajectory: The path of a projectile (a parabola if air resistance is ignored).
Time of flight: Total time a projectile spends in the air.
Range: Horizontal distance travelled by a projectile.
Angular speed (ω): Angle turned per second, in rad/s; ω = 2π/T.
Period (T): Time for one complete revolution or oscillation.
Frequency (f): Number of oscillations or revolutions per second; f = 1/T; unit hertz (Hz).
Centripetal acceleration: Acceleration towards the centre of a circle, v²/r.
Centripetal force: The net force towards the centre needed for circular motion.
Simple harmonic motion: Oscillation in which acceleration ∝ −displacement.
Amplitude: Maximum displacement from equilibrium.
Equilibrium position: The rest position, where the restoring force is zero.

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