A small ball moves so that x = 3t and y = 4t - 5t², in metres, with t in seconds…

Step 1. Remove t. $t = \dfrac{x}{3}$, so $y = 4 \times \dfrac{x}{3} - 5 \times \dfrac{x^2}{9} = \dfrac{4}{3}x - \dfrac{5}{9}x^2$.

Step 2. The curve. This has the form $y = ax - bx^2$: a parabola that opens downwards, like the path of a thrown ball.

Step 3. Back to y = 0. $x\left(\dfrac{4}{3} - \dfrac{5}{9}x\right) = 0$, so $x = 0$ or $x = \dfrac{4}{3} \times \dfrac{9}{5} = 2.4\ \text{m}$.

Step 4. Highest point. Halfway, at $x = 1.2\ \text{m}$: $y = \dfrac{4}{3}(1.2) - \dfrac{5}{9}(1.44) = 1.6 - 0.8 = 0.8\ \text{m}$.