B

Step 1: For a projectile kicked at an angle \( \theta \) with the horizontal, the maximum horizontal range \( R \) is given by \( R = \frac{v^2 \sin(2\theta)}{g} \).
Step 2: The greatest height \( H \) attained by the projectile is given by \( H = \frac{v^2 \sin^2(\theta)}{2g} \).
Step 3: To find the relation between \( H \) and \( R \), we know that at maximum range \( \theta = 45^{\circ} \), therefore \( \sin(2\theta) = 1 \) and \( \sin(\theta) = \frac{\sqrt{2}}{2} \).
Step 4: Substituting \( \theta = 45^{\circ} \) into the equation for height gives: \( H = \frac{v^2 \left(\frac{\sqrt{2}}{2}\right)^2}{2g} = \frac{v^2}{4g} \).
Step 5: We can see from the equations that \( H = \frac{R}{2} \) when rearranging gives the answer: \( H = \frac{R}{2} \times 2 = R \) so this leads to a conclusion that \( H = 2R \).
Therefore, the correct option is B.