By Definition of Kinetic Energy, it is valid:
[tex]E_c= \frac{mv^2}{2} \\ E_c= \frac{1\times 30^2}{2} \\ \boxed {E_c=450~J}[/tex]
Using the Torricelli's Equation in vertical direction on the maximum height, remember that the final velocity at this point is zero, we have:
[tex]v^2=v_{o}^2+2a\Delta S \\ 0=30^2+2(-10)H \\ H= \frac{900}{20} \\ \boxed {H=45~m} [/tex]
If you notice any mistake in my english, please let me know, because my english level is still low.
