We assume no air resistance, so mechanical energy is conserved. Let the ground be the zero potential energy level ($U_g = 0$).

[Q1] For a vertical launch, the velocity at the highest point is zero. By conservation of energy, the initial kinetic energy is fully converted to potential energy at the peak.

$$U_{g,max} = K_i = \frac{1}{2}mv_i^2$$ $$U_{g,max} = \frac{1}{2}(10 \text{ kg})(500 \text{ m/s})^2$$

[Q2] For an angled launch, the velocity at the highest point is the constant horizontal component of the initial velocity, $v_{top} = v_x = v_i \cos\theta$. The kinetic energy at the highest point is:

$$K_{top} = \frac{1}{2}mv_{top}^2 = \frac{1}{2}m(v_i \cos\theta)^2$$ $$K_{top} = \frac{1}{2}(10 \text{ kg})(500 \text{ m/s} \cdot \cos 60^\circ)^2$$

By conservation of energy, $E_i = E_{top}$, so $K_i = K_{top} + U_{g,max}$. The potential energy at the highest point is:

$$U_{g,max} = K_i - K_{top} = \frac{1}{2}mv_i^2 - \frac{1}{2}m(v_i \cos\theta)^2 = \frac{1}{2}mv_i^2\sin^2\theta$$ $$U_{g,max} = \frac{1}{2}(10 \text{ kg})(500 \text{ m/s} \cdot \sin 60^\circ)^2$$