Projectile motion: WACE Year 12 Physics Unit 3

What this dot point is asking

WACE wants you to treat projectile motion as two simultaneous one-dimensional problems joined only by time. The defining idea is independence: what happens horizontally does not affect what happens vertically. A horizontally thrown ball and a dropped ball hit the ground together because both share the same downward acceleration and the same vertical start.

Resolving the launch velocity

For a launch speed $u$ at angle $\theta$ above the horizontal,

The horizontal component stays fixed for the whole flight. The vertical component decreases on the way up, reaches zero at the peak, then grows downward.

The two sets of equations

Horizontally, with no acceleration,

Vertically, taking up as positive with $a=-g$,

Time is the bridge: you usually solve the vertical equation for $t$, then feed that time into the horizontal equation to get range.

Key flight quantities

At maximum height $v_y=0$, so the time to the peak is $t_{\text{up}}=u_y/g$. For a launch and landing at the same height, the total flight time is $2u_y/g$ and the horizontal range is

This shows range is greatest at $45^\circ$ and that complementary angles (for example $30^\circ$ and $60^\circ$) give the same range. The speed at any instant is $v=\sqrt{v_x^2+v_y^2}$ and its direction is $\tan^{-1}(v_y/v_x)$.

For a projectile launched from a height (such as off a cliff or table), the flight is no longer symmetric, so you cannot simply double the time to the peak. Instead, set up the vertical displacement equation with the correct sign for the landing position below the launch point and solve the resulting quadratic for $t$. Once you have the flight time, the horizontal range follows from $x=u_x t$ as usual. A horizontally launched projectile is just the special case with $u_y=0$, where the whole launch speed is horizontal and the object begins falling immediately.

Reading the trajectory

The path is a parabola. At the top the velocity is purely horizontal (equal to $u_x$), never zero. The vertical speed at landing equals the launch vertical speed for symmetric flight. Air resistance, which WACE ignores unless told otherwise, would shorten the range and make the descent steeper than the ascent.

The non-symmetric launch in detail

The cliff or table launch is where most marks are lost, so treat it carefully. Choose up as positive and place the origin at the launch point. The vertical displacement at landing is then negative (the ground is below the launch), which is the sign that makes the quadratic $y=u_y t-\tfrac{1}{2}gt^2$ produce a sensible positive time. Solving that quadratic with the formula gives two roots; discard the negative one as unphysical. Only after you have the flight time should you compute the horizontal range $x=u_x t$ and the landing velocity components $v_x=u_x$ and $v_y=u_y-gt$. Because the descent is longer than the ascent for an upward launch off a height, the landing speed exceeds the launch speed, which is a useful sanity check.

Setting up cleanly

Always declare a positive direction and list the known quantities separately for each axis before substituting. Mixing a horizontal speed into a vertical equation is the fastest way to lose marks. Carry one extra figure through working and round only at the end.