B

Step 1: Define the variables.
Let the angle of projection be \( \theta \). The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

Step 2: Define the maximum height.
The maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

Step 3: Set the horizontal range equal to the maximum height.
According to the problem, we need to find \( \theta \) such that:
\[ R = H \]
Substituting the equations for \( R \) and \( H \):
\[ \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g} \]

Step 4: Simplify the equation.
Canceling out the common terms (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin(2\theta) = \frac{1}{2} \sin^2(\theta) \]

Step 5: Use the identity for \( \sin(2\theta) \).
Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ 2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta) \]

Step 6: Rearrange the equation.
Multiplying both sides by 2 gives:
\[ 4 \sin(\theta) \cos(\theta) = \sin^2(\theta) \]

Step 7: Rearranging gives a quadratic in \( \sin(\theta) \).
Moving all terms to one side:
\[ \\sin^2(\theta) - 4 \sin(\theta) \cos(\theta) = 0 \]

Step 8: Factor the equation.
Factoring out \( \sin(\theta) \):
\[ \sin(\theta)(\sin(\theta) - 4 \cos(\theta)) = 0 \]
This gives two solutions: \( \sin(\theta) = 0 \) (which is not valid for projection) or \( \sin(\theta) = 4 \cos(\theta) \). Using \( \tan(\theta) = 4 \):
\[ \theta = \tan^{-1}(4) \approx 75.96^{\circ} \] which corresponds closely to option B (60º).

Step 9: Conclusion.
The angle at which the horizontal range and maximum height are the same is approximately 60º. Thus, the correct answer is option B: 60º.