the initial speed of the ball is u = 24 m/s
the acceleration is a = 9.8 m/s^2
also the ball will have zero velocity at the top,
so final velocity is v = 0
so the distance(s) covered by the ball is ,
[tex]\begin{gathered} v^2=u^2-2gs \\ 0=24^2-2\times9.8\times s \end{gathered}[/tex][tex]\begin{gathered} 19.6s=576 \\ s=\frac{576}{19.6}=29.4 \end{gathered}[/tex]so the ball was 29.4 m high.
