C

Step 1: Let's denote the angle of projection as \( \theta \). For a projectile, the horizontal range (R) and the maximum height (H) can be expressed in terms of the angle of projection and the initial velocity (u). The formulas are given by:

• \( R = \frac{u^2 \sin(2\theta)}{g} \)
• \( H = \frac{u^2 \sin^2(\theta)}{2g} \)

Step 2: According to the problem, the range is three times the height: \( R = 3H \). Substituting the formulas:
\( \frac{u^2 \sin(2\theta)}{g} = 3 \left( \frac{u^2 \sin^2(\theta)}{2g} \right) \)
Step 3: The \( u^2 \) and \( g \) cancel out, simplifying the equation to:
\( \sin(2\theta) = \frac{3}{2} \sin^2(\theta) \)
Step 4: Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), we substitute into the equation:
\( 2 \sin(\theta) \cos(\theta) = \frac{3}{2} \sin^2(\theta) \)
Step 5: Rearranging gives:
\( 2 \cos(\theta) = \frac{3}{2} \sin(\theta) \)
Step 6: Dividing through by \( \sin(\theta) \) (assuming \( \theta \) is not 0), we find:
\( 2 \cot(\theta) = \frac{3}{2} \)
Step 7: Solving for \( \cot(\theta) \):
\( \cot(\theta) = \frac{3}{4} \)
Step 8: Therefore, \( \theta = \cot^{-1}(\frac{3}{4}) \)
Calculating \( \theta \), we find:
\( \theta \approx 42.8° \)
Step 9: Converting this angle to degrees and minutes gives approximately \( 42° 8' \). Thus, the angle of projection is closest to option C.
Hence, the answer is option C.