We use the principle of conservation of mechanical energy, assuming no air resistance. Let the launch point be the reference level for gravitational potential energy ($h=0$). The initial total energy is purely kinetic:

$$E_i = \frac{1}{2}mv_0^2$$

At a height $h$ above the launch point, the projectile has speed $v$. The total energy is the sum of its kinetic and potential energy:

$$E_f = \frac{1}{2}mv^2 + mgh$$

By conservation of energy, $E_i = E_f$:

$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$$

Solving for the speed $v$:

$$v^2 = v_0^2 - 2gh$$ $$v = \sqrt{v_0^2 - 2gh}$$

The speed $v$ at height $h$ depends only on the initial speed $v_0$ and the height $h$, not on the launch angle.