Projectile motion (SACE Stage 2 Physics)

What this dot point is asking

You must model the flight of a projectile by resolving the launch velocity into horizontal and vertical components, treating each component independently, and using the shared time of flight to connect them. From this you predict time of flight, range, maximum height and the velocity at any instant.

The key idea: independence of the two directions

Once a projectile leaves the hand, ramp or muzzle, the only force on it is gravity (air resistance is ignored at this level). Gravity acts vertically, so:

• Horizontal: no force, so no acceleration. The horizontal velocity $u_x$ stays constant for the whole flight.
• Vertical: constant acceleration $g$ downward, so the vertical velocity changes uniformly.

The two motions share one quantity only: time. The projectile spends the same time $t$ in the air for both its horizontal and vertical travel. This is what lets you solve one direction for $t$ and feed it into the other.

The equations you use

Horizontal (constant velocity):

Vertical (constant acceleration, taking up as positive, $a = -g$):

Be consistent with your sign convention. Many students take up as positive so $g$ enters as $-9.80\ \text{m s}^{-2}$; others take down as positive for a falling object. Either works as long as you keep it consistent across the whole problem. SACE marking schemes accept any consistent convention, so state it clearly at the top of your working.

Maximum height, time of flight and range

At maximum height the vertical velocity is momentarily zero ($v_y = 0$), but the horizontal velocity is unchanged. Setting $v_y = 0$ gives the time to the top, and doubling it (for a projectile landing at launch height) gives the total time of flight.

• Time to peak: $t_\text{up} = \dfrac{u_y}{g}$
• Maximum height above launch: $H = \dfrac{u_y^2}{2g}$
• Range on level ground: $R = u_x \, t_\text{flight}$

When the launch and landing heights differ (a cliff, a raised platform, a ball struck off a tee into a valley), the symmetry shortcut fails. Instead, write the vertical displacement equation $y = u_y t - \tfrac{1}{2} g t^2$ with $y$ equal to the (signed) drop and solve the resulting quadratic for $t$, taking the positive root.

Velocity at any point

The velocity at any instant is the vector sum of the (constant) horizontal component and the (changing) vertical component:

where $\alpha$ is the angle of the velocity above or below the horizontal. The speed is least at the top of the path (where $v_y = 0$ and $v = v_x$) and greatest at the lowest point of the trajectory.

The trajectory is a parabola

Eliminating $t$ between $x = u_x t$ and $y = u_y t - \tfrac{1}{2}gt^2$ gives $y$ as a quadratic in $x$, a parabola. This is why projectile paths are symmetric about the highest point when the launch and landing heights are equal. It also explains why two different launch angles that add to $90^\circ$ (for example $30^\circ$ and $60^\circ$) give the same range on level ground: the product $u_x u_y \propto \sin 2\theta$, which is symmetric about $45^\circ$.

How SACE assesses this

SACE Stage 2 projectile questions almost always supply pre-resolved components or a diagram, then ask you to "show that" a given value follows, or to decide whether an object clears an obstacle. Lay out the horizontal and vertical motions in separate columns, quote the equation before substituting, and finish with a sentence that answers the question in context. The "show that" parts reward working that visibly arrives at the printed value, so do not skip the substitution line.