Step 1: Write the range as a function of the launch angle.
For a fixed launch speed \( v_0 \), the horizontal range is
\[
R(\theta) = \frac{v_0^2}{g}\sin(2\theta)
\]
Step 2: Differentiate and set the slope to zero.
To find where \( R \) is largest, take the derivative with respect to \( \theta \) and set it to zero:
\[
\frac{dR}{d\theta} = \frac{v_0^2}{g}\cdot 2\cos(2\theta) = 0 \quad \Rightarrow \quad \cos(2\theta) = 0
\]
Step 3: Solve for \( \theta \) and confirm it is a maximum.
\[
2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ
\]
The second derivative is \( \frac{d^2R}{d\theta^2} = -\frac{4v_0^2}{g}\sin(2\theta) \), which is negative at \( \theta = 45^\circ \) since \( \sin(90^\circ) = 1 \gt 0 \), confirming this is indeed a maximum.
So the range is maximum at \( \theta = \boxed{45^\circ} \), matching option (B).