Substitute t = 10, and t = 20, to the given equation and we get
[tex]\begin{gathered} \text{If }t=10 \\ s=120t-6t^2 \\ s=120(10)-6(10)^2 \\ s=1200-6(100) \\ s=1200-600 \\ s=600 \\ \\ \text{If }t=20 \\ s=120t-6t^{2} \\ s=120(20)-6(20)^2 \\ s=2400-6(400)^2 \\ s=2400-2400 \\ s=0 \end{gathered}[/tex]We therefore have t = 10 as the time it takes to reach the highest point, with the rock reaching 600m.
Therefore, we choose second option.