Projectile and circular motion - TCE Physics (Tasmania)

Syllabus dot point

“Analyse projectile motion and uniform circular motion using vectors and Newton's laws.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

A projectile has independent horizontal (constant velocity) and vertical (constant acceleration $g$ downward) motions; an object in uniform circular motion moves at constant speed but accelerates toward the centre because a net centripetal force continually changes its direction.

What this dot point is asking

This dot point combines the kinematics of straight-line motion with Newton's laws to handle two of the most common exam scenarios: a launched projectile and an object turning in a circle.

Projectile motion

A projectile is any object moving only under gravity after launch, with air resistance ignored. The key idea is that the horizontal and vertical components of motion are completely independent and share only the time of flight.

Horizontally there is no force, so velocity is constant:

x = u_x t

where $u_x = u\cos\theta$ is the horizontal launch component.

Vertically the only acceleration is gravity, $g = 9.8\ \text{m s}^{-2}$ downward, so the constant-acceleration equations apply:

v_y = u_y - g t \qquad y = u_y t - \tfrac{1}{2} g t^2

with $u_y = u\sin\theta$ . At the highest point $v_y = 0$ , which lets you find the time to the top and the maximum height. The trajectory is a parabola, and for level ground the path is symmetric: the launch speed equals the landing speed, and the launch angle equals the landing angle below the horizontal.

To find the resultant velocity at any instant, combine the components:

v = \sqrt{v_x^2 + v_y^2} \qquad \tan\phi = \frac{v_y}{v_x}.

Worked example: horizontal launch from a cliff

A ball leaves the top of a $45\ \text{m}$ high cliff horizontally at $20\ \text{m s}^{-1}$ . Find how long it takes to hit the ground and how far from the base of the cliff it lands.

Step 1: Use vertical motion to find the time of flight

Because the ball is launched horizontally, its initial vertical velocity is zero ( $u_y = 0$ ). Gravity then accelerates it downward, so the vertical displacement follows $y = \tfrac{1}{2} g t^2$ . Setting $y$ equal to the cliff height of $45\ \text{m}$ and solving for $t$ gives the time to reach the ground, since that is the only unknown:

y = \tfrac{1}{2} g t^2 \Rightarrow 45 = \tfrac{1}{2}(9.8)t^2, \quad t^2 = \frac{90}{9.8} = 9.18, \quad t = 3.03\ \text{s}.

Step 2: Use horizontal motion for the range

Horizontally there is no force, so the ball travels at constant speed $u_x = 20\ \text{m s}^{-1}$ throughout the flight. Multiplying the constant horizontal speed by the flight time found in Step 1 gives the horizontal range:

x = u_x t = (20)(3.03) = 60.6\ \text{m}.

Final answer: The ball lands after about $3.0\ \text{s}$ , $61\ \text{m}$ from the base of the cliff.

Uniform circular motion

An object moving in a circle at constant speed is still accelerating because its velocity direction changes continually. This acceleration points toward the centre and is the centripetal acceleration:

a_c = \frac{v^2}{r}

By Newton's second law a net force must produce it, the centripetal force:

F_c = \frac{m v^2}{r}

This is not a new kind of force. It is whatever real force (tension, friction, gravity, the normal force, or a magnetic force) happens to point toward the centre. If you know the period $T$ for one revolution, the speed is the circumference divided by the period:

v = \frac{2\pi r}{T}.

The centripetal force does no work on the object because it is always perpendicular to the velocity, which is why the speed stays constant even though the object accelerates.

Conical pendulums and banked turns

A favourite TCE scenario is an object swung in a horizontal circle on a string at an angle, or a vehicle on a banked or flat curve. The method is always the same: draw the forces, resolve into vertical and horizontal, set the vertical components to balance (no vertical acceleration) and the net horizontal force equal to $\dfrac{mv^2}{r}$ . For a string at angle $\theta$ from the vertical, dividing the horizontal equation by the vertical gives $\tan\theta = \dfrac{v^2}{rg}$ , which neatly eliminates the tension and the mass.

Centre-directed acceleration

In uniform circular motion the speed is constant but the velocity is not, so there is always a non-zero acceleration directed toward the centre of the circle.

The link between the two halves of this dot point is the role of gravity. A projectile near the surface follows a parabola because gravity is effectively constant and vertical; a satellite follows a circle (or ellipse) because gravity always points toward the central mass and supplies the centripetal force. Newton's famous cannonball thought-experiment connects them: fire a projectile fast enough and its parabolic fall curves around the whole planet, becoming an orbit.

Centrifugal force does not act on the object

Do not add an outward "centrifugal force" to the object in your force diagram. The only real force is the inward (centripetal) force. The outward feeling you experience comes from your inertia, your tendency to travel in a straight line, not from a real outward force on the body moving in the circle.

When you analyse circular motion in the exam, first identify which real force provides the centripetal force, then set that force equal to $\dfrac{mv^2}{r}$ and solve. For a car on a flat curve it is friction; for a satellite it is gravity; for a charge in a magnetic field it is the magnetic force. Keeping the source force separate from the centripetal requirement is the clearest way to avoid confusion.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A parent swings a child of mass

25\ \text{kg}

in a horizontal circle by the arms, radius

1.3\ \text{m}

, with the arms held at

20^\circ

above the horizontal. Calculate the magnitude of the force on the child's arms and the tangential speed of the child.

Show worked answer →

Two forces act on the child: weight $mg$ (down) and the arm force $T$ along the arms ( $20^\circ$ above the horizontal). The vertical component of $T$ balances weight; the horizontal component provides the centripetal force.

Vertical: $T\sin 20^\circ = mg$ , so $T = \dfrac{mg}{\sin 20^\circ} = \dfrac{(25)(9.81)}{0.342} = 717\ \text{N}$ .

Horizontal (centripetal): $T\cos 20^\circ = \dfrac{mv^2}{r}$ , so $\dfrac{mv^2}{r} = (717)(0.940) = 674\ \text{N}$ , giving $v^2 = \dfrac{(674)(1.3)}{25} = 35.0$ and $v = 5.92\ \text{m s}^{-1}$ .

The tangential speed is about $5.9\ \text{m s}^{-1}$ and the arm force about $717\ \text{N}$ . Markers want the vertical balance to find $T$ , then the horizontal component as the centripetal force.

Practice3 marks

A cricket six has a range of

120\ \text{m}

, struck at

40^\circ

from ground level and landing at ground level. Using

u\cos 40^\circ = \dfrac{120}{t}

and

u\sin 40^\circ = 4.91t

, calculate the initial speed

u

of the ball.

Show worked answer →

Treat the projectile as independent horizontal and vertical motion. The two relations come from the range (horizontal) and the time of flight (the ball returns to the same height after time $t$ when $u\sin 40^\circ = \tfrac12 gt = 4.91t$ ).

From the first: $t = \dfrac{120}{u\cos 40^\circ}$ . Substitute into the second:

u\sin 40^\circ = \frac{(4.91)(120)}{u\cos 40^\circ},

u^2 \sin 40^\circ \cos 40^\circ = (4.91)(120) = 589.2

u^2 = \frac{589.2}{\sin 40^\circ \cos 40^\circ} = \frac{589.2}{0.4924} = 1196.6, \quad u = 34.6\ \text{m s}^{-1}.

The ball leaves the bat at about $35\ \text{m s}^{-1}$ . Markers want elimination of $t$ between the two equations.

Practice3 marks

A golf ball is hit at

140\ \text{km h}^{-1}

40^\circ

to the horizontal on flat ground (

g = 9.81\ \text{m s}^{-2}

). Calculate the time the ball is above the ground, ignoring air resistance.

Show worked answer →

Convert the speed: $140\ \text{km h}^{-1} = \dfrac{140}{3.6} = 38.9\ \text{m s}^{-1}$ .

Vertical launch component: $u_y = 38.9\sin 40^\circ = (38.9)(0.643) = 25.0\ \text{m s}^{-1}$ upwards.

Time to the top (where $v_y = 0$ ): $t_\text{up} = \dfrac{u_y}{g} = \dfrac{25.0}{9.81} = 2.55\ \text{s}$ .
Total flight $= 2 t_\text{up} = 5.10\ \text{s}$ .

The ball is in the air for about $5.1\ \text{s}$ . Markers want the vertical component, the time to apex, then doubling for the full symmetric flight.

Practise this

Exam-style questions 3 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC