C

Given: A body is projected with an initial velocity u. The maximum height (H) attained is half of its horizontal range (R). We need to find the expression for the range R.

Step 1: Understand the motion equations.
For a projectile launched at an angle θ with initial velocity u, the maximum height is given by the formula:
$$ H = \frac{u^2 \sin^2(\theta)}{2g} $$
The horizontal range is given by:
$$ R = \frac{u^2 \sin(2\theta)}{g} $$

Step 2: Set up the relation between maximum height and range.
According to the problem, we have:
$$ H = \frac{1}{2} R $$
Substituting the formulas we derived for H and R:
$$ \frac{u^2 \sin^2(\theta)}{2g} = \frac{1}{2} \left( \frac{u^2 \sin(2\theta)}{g} \right) $$

Step 3: Simplify the equation.
Multiplying both sides by 2g results in:
$$ u^2 \sin^2(\theta) = \frac{u^2 \sin(2\theta)}{1} $$
Using the identity sin(2θ) = 2sin(θ)cos(θ):
$$ u^2 \sin^2(\theta) = u^2 \cdot 2 \sin(\theta) \cos(\theta) $$
Assuming u is not equal to zero, we can divide by u^2:
$$ \sin^2(\theta) = 2 \sin(\theta) \cos(\theta) $$

Step 4: Rearranging the trigonometric identity.
This can be rewritten as:
$$ \sin^2(\theta) - 2 \sin(\theta) \cos(\theta) = 0 $$
Factoring gives:
$$ \sin(\theta)(\sin(\theta) - 2 \cos(\theta)) = 0 $$
This results in:
1) $\sin(\theta) = 0$, leads to no projection, or
2) $\sin(\theta) = 2 \cos(\theta)$ leading to $\tan(\theta) = 2$ and hence $\theta = \tan^{-1}(2)$.

Step 5: Calculate the range.
Substituting back to find range:
$$ R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \cdot \frac{2 \cdot 2}{\sqrt{5}}}{g} = \frac{4u^2 /5}{g} = \frac{4u^2}{5g} $$

Final Step: Conclusion.
This shows that the range R must be:
$$ R = \frac{4u^2}{5g} $$
Thus, the correct answer is option C.