A ball is projected from the ground with a speed $$15 \text{ m s}^{-1}$$ at an angle $$\theta$$ with horizontal so that its range and maximum height are equal, then $$\tan \theta$$ will be equal to

A ball is projected with speed $$15 \text{ m s}^{-1}$$ at angle $$\theta$$ such that range equals maximum height.

To determine the required angle, we first write the expressions for the range and the maximum height.

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

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

Equating the range and the maximum height gives

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

Since $$u^2 \sin\theta / g \neq 0$$, cancelling common factors yields

$$2\cos\theta = \frac{\sin\theta}{2}$$

$$4\cos\theta = \sin\theta$$

$$\tan\theta = 4$$

Therefore, the correct answer is Option D: $$4$$.