A ball is launched at 10 m/s, at 45° above level ground…

Step 1. The numbers. $\tan 45^\circ = 1$, and $2u^2\cos^2 45^\circ = 2 \times 100 \times 0.5 = 100$.

Step 2. The path. $y = x - \dfrac{9.8\,x^2}{100} = x - 0.098\,x^2$.

Step 3. At 5 m. $y = 5\ \text{m} - 0.098\ \text{m}^{-1} \times (5\ \text{m})^2 = 5\ \text{m} - 2.45\ \text{m} = 2.55\ \text{m}$.

Step 4. Landing. $x(1 - 0.098x) = 0$, so $x = \dfrac{1}{0.098\ \text{m}^{-1}} = 10.2\ \text{m}$.