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

<p>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:</p> <p>Range:</p> <p>$R = \frac{u^2 \sin(2\theta)}{g}$</p> <p>Maximum height:</p> <p>$H = \frac{u^2 \sin^2(\theta)}{2g}$</p> <p>We are given that the range and maximum height are equal. So, we set these equations equal to each other:</p> <p>$\frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g}$</p> <p>We can cancel out the common terms $u^2$ and $g$ on both sides:</p> <p>$\sin(2\theta) = \frac{1}{2} \sin^2(\theta)$</p> <p>Using the double angle identity for sine, $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we substitute it into the equation:</p> <p>$2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta)$</p> <p>We can simplify this by dividing both sides by $\sin(\theta)$ (assuming $\theta \neq 0$):</p> <p>$2 \cos(\theta) = \frac{1}{2} \sin(\theta)$</p> <p>Rearranging to get all terms on one side gives us:</p> <p>$4 \cos(\theta) = \sin(\theta)$</p> <p>Dividing both sides by $\cos(\theta)$, we get:</p> <p>$4 = \tan(\theta)$</p> <p>So, we find that:</p> <p>$\tan(\theta) = 4$</p> <p>Thus, the correct answer is:</p> <p>Option D: 4</p>