For a projectile projected with speed $u$ at angle $\theta$, the horizontal range is $R=\frac{u^2\sin 2\theta}{g}.$

The first projectile has maximum range. This happens when $\sin 2\theta=1 \Rightarrow 2\theta=90^\circ \Rightarrow \theta=45^\circ.$ So the maximum range is $R_{\max}=\frac{u^2}{g}.$

For the second projectile, the range is half of this maximum range: $R_2=\frac{1}{2}R_{\max}=\frac{1}{2}\cdot \frac{u^2}{g}.$

Using the range formula for the second projectile with the same speed $u$: $\frac{u^2\sin 2\theta}{g}=\frac{1}{2}\cdot \frac{u^2}{g}.$ Cancel $\frac{u^2}{g}$ from both sides: $\sin 2\theta=\frac{1}{2}.$

Now solve: $2\theta=30^\circ \quad \text{or} \quad 150^\circ.$ Hence, $\theta=15^\circ \quad \text{or} \quad 75^\circ.$

Therefore, the possible projection angles are: $\boxed{15^\circ \text{ or } 75^\circ}.$