To solve this problem, we will use the equations for the range and maximum height of a projectile. The range $$R$$ and maximum height $$H$$ of a projectile launched with speed $$u$$ at an angle $$\theta$$ can be expressed as follows:

Range:

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

Maximum height:

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

We are given that the range and maximum height are equal. So, we set these equations equal to each other:

$$\frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g}$$

We can cancel out the common terms $$u^2$$ and $$g$$ on both sides:

$$\sin(2\theta) = \frac{1}{2} \sin^2(\theta)$$

Using the double angle identity for sine, $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$$, we substitute it into the equation:

$$2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta)$$

We can simplify this by dividing both sides by $$\sin(\theta)$$ (assuming $$\theta \neq 0$$):

$$2 \cos(\theta) = \frac{1}{2} \sin(\theta)$$

Rearranging to get all terms on one side gives us:

$$4 \cos(\theta) = \sin(\theta)$$

Dividing both sides by $$\cos(\theta)$$, we get:

$$4 = \tan(\theta)$$

So, we find that:

$$\tan(\theta) = 4$$

Thus, the correct answer is:

Option D: 4