Explanation:
The final velocity at the cannon ball's maximum height is zero ([tex]v_y = 0[/tex]). We can use the equation
[tex]v_y^2 = 0 = v_{0y}^2 - 2gy_{max}[/tex]
[tex]\Rightarrow y_{max} = \dfrac{v_{0y}^2}{2g}[/tex]
[tex]y_{max} = \dfrac{(25\:\text{m/s})^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:\:=31.9\:\text{m}[/tex]
