Answer :
Given data:
Initial velocity of the rock;
[tex]u=600\text{ m/s}[/tex]Launch angle;
[tex]\theta=60\degree[/tex]The maximum height the rock will go is given as,
[tex]H=\frac{u^2\sin ^2\theta}{2g}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} H=\frac{(600\text{ m/s})^2\times\sin ^2(60\degree)}{2\times(9.8\text{ m/s}^2)} \\ \approx13775.51\text{ m} \end{gathered}[/tex]Therefore, the rock will attain a maximum height of 13775.51 m.
The horizontal range is given as,
[tex]R=\frac{u^2\sin (2\theta)}{g}[/tex]Substituting all known values,
[tex]\begin{gathered} R=\frac{(600\text{ m/s})^2\times\sin (2\times60\degree)}{(9.8\text{ m/s}^2)} \\ \approx31813.18\text{ m} \end{gathered}[/tex]Therefore, the rock will land 31813.18 m away from the launch site.