How to solve projectile motion problems
Treat the two directions separately, connect them through time, and avoid the sign error that ruins the vertical calculation.
What the examiner is testing
Whether you keep the two directions apart. Nearly every lost mark comes from mixing horizontal and vertical quantities in the same equation.
Two facts drive everything:
- Horizontally: no force (ignoring air resistance), so velocity is constant.
- Vertically: acceleration is \( g = 9.81 \text{ m/s}^2 \) downwards, always.
Time is the bridge. It is the only quantity shared by both directions, and it is how nearly every one of these questions is solved.
The method
- Resolve the initial velocity: \( u_x = u\cos\theta \), \( u_y = u\sin\theta \).
- Choose a positive direction for the vertical and state it.
- Use the vertical motion to find the time.
- Feed that time into the horizontal, where \( s_x = u_x t \).
Worked example
A ball is kicked at 20 m/s at 30° above horizontal from level ground. Find the time of flight and the horizontal range.
Step 1 — resolve.
$$ u_x = 20\cos 30° = 17.3 \text{ m/s} $$
$$ u_y = 20\sin 30° = 10.0 \text{ m/s} $$
Step 2 — sign convention. Take upwards as positive, so \( a = -9.81 \text{ m/s}^2 \).
Step 3 — vertical, to find time. It lands at the same height it left, so the vertical displacement over the whole flight is \( s_y = 0 \):
$$ s_y = u_y t + \tfrac{1}{2}at^2 \quad \Rightarrow \quad 0 = 10.0t - 4.905t^2 $$
Factorising, \( t(10.0 - 4.905t) = 0 \), so \( t = 0 \) or:
$$ t = \frac{10.0}{4.905} = 2.04 \text{ s} $$
The \( t = 0 \) root is the moment of launch. Discard it — but recognise it, because it is a sign the algebra is right.
Step 4 — horizontal, using that time.
$$ s_x = u_x t = 17.3 \times 2.04 = 35.3 \text{ m} $$
Answer: time of flight 2.04 s, range 35.3 m.
Check: the time to the top should be half the flight, \( 1.02 \text{ s} \). From \( v = u + at \) with \( v = 0 \): \( t = 10.0/9.81 = 1.02 \text{ s} \). Consistent.
The three mistakes that lose marks
1. Using the resultant speed in a vertical equation. Only \( u_y \) goes into vertical suvat. Resolve first, every time.
2. Sign confusion. Pick a positive direction, write it down, and stay with it. If up is positive then \( g \) is negative — mixing these gives a negative time, which is the giveaway.
3. Assuming the flight is symmetric when it is not. Equal launch and landing heights make it symmetric. Launch from a cliff, and it is not — you must solve the quadratic properly rather than doubling the time to the peak.
30-second recap
Resolve into components. Vertical gives you time; horizontal uses it. State your sign convention before writing any equation. At the top, vertical velocity is zero. Check by confirming the time to the peak is half the flight time — when the heights match.