TASPhysicsSyllabus dot point

How do objects move when launched or turned by a force?

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

How to analyse projectiles by separating horizontal and vertical motion, and how a centripetal force keeps objects in uniform circular motion.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

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 standard 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 with Pythagoras and trigonometry:

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

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 called 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.

When you analyse circular motion in the exam, first identify which real force provides the centripetal force, then set that force equal to $\frac{m v^2}{r}$ 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 physics of the source force separate from the centripetal requirement is the clearest way to avoid confusion.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC4 marksA parent is swinging their child in a circle by the arms. The child has a mass of 25 kg and the radius of the circle is 1.3 m. The child is supported at a constant angle of 20 degrees to the horizontal. Calculate the magnitude of the force the parent exerts on the child's arms, and the tangential speed of the child.

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

Vertical: T sin20 = mg, so T = mg / sin20 = 25 x 9.81 / 0.342 = 717 N. This is the force on the arms.

Horizontal (centripetal): T cos20 = m v^2 / r.
T cos20 = 717 x 0.940 = 674 N.
So m v^2 / r = 674, giving v^2 = 674 x 1.3 / 25 = 35.0, and v = 5.92 m s-1.

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

2024 TASC3 marksDuring a game of cricket a player hits a six with a range of 120 m, struck at 40 degrees from ground height and landing at ground height. Using the relations u cos40 = 120 / t and u sin40 = 4.91 t, calculate the initial speed u of the ball.

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Treat the projectile as independent horizontal and vertical motion. The two given relations come from the range (horizontal) and the time of flight (vertical, since the ball returns to the same height after time t when u sin40 = g t / 2 = 4.91 t).

From the first relation: t = 120 / (u cos40).
Substitute into the second: u sin40 = 4.91 x 120 / (u cos40).
So u^2 sin40 cos40 = 4.91 x 120 = 589.2.
u^2 = 589.2 / (sin40 cos40) = 589.2 / (0.643 x 0.766) = 589.2 / 0.4924 = 1196.6.
u = 34.6 m s-1.

The ball leaves the bat at about 35 m s-1. Markers want elimination of t between the two equations and use of sin40 cos40.

2019 TASC3 marksA golf ball is hit on Earth (g = 9.81 m s-2) at 140 km/h at 40 degrees to the horizontal. Calculate the time the ball is above the ground. Ignore air resistance and assume flat ground.

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Convert the speed: 140 km/h = 140 / 3.6 = 38.9 m s-1.

Vertical launch component: u_y = 38.9 sin40 = 38.9 x 0.643 = 25.0 m s-1 (upwards).

The ball returns to the same height, so use v = u + at over the whole flight with the vertical velocity reversing symmetrically. Time to the top (where v_y = 0): t_up = u_y / g = 25.0 / 9.81 = 2.55 s.
Total time of flight = 2 x t_up = 5.10 s.

The ball is in the air for about 5.1 s. Markers want the vertical component, time to apex, then doubling for the full flight.

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