In a computer game, a ball is at (2, 1) m at t = 0.

Step 1. Along x. $x = 2 + 1 \times 3 = 5\ \text{m}$, and $v_x$ stays $1\ \text{m/s}$.

Step 2. Along y. $y = 1 + 3 \times 3 + \tfrac{1}{2}(-2)(3^2) = 1\ \text{m}$, and $v_y = 3 - 2 \times 3 = -3\ \text{m/s}$.

Step 3. Speed. $\sqrt{1^2 + (-3)^2} = 3.16\ \text{m/s}$.

Step 4. Highest point. $v_y = 3 - 2t = 0$ at $t = 1.5\ \text{s}$.