Step 1: Picture the projectile.
A ball is thrown from the ground at speed $v$ and angle $\theta$. It goes up, reaches a top point, and comes back down. The total horizontal distance it covers is the range $R$. The highest point it reaches is the maximum height $H$. We are told these two are equal.
Step 2: Write the height formula.
The maximum height of a projectile is \[ H = \frac{v^2 \sin^2\theta}{2g} \] This comes from the vertical motion, where the upward speed slowly becomes zero at the top.
Step 3: Write the range formula.
The horizontal range is \[ R = \frac{v^2 \sin 2\theta}{g} \] This tells how far the ball lands from the start.
Step 4: Set them equal.
Since $R = H$, we put the two formulas equal: \[ \frac{v^2 \sin^2\theta}{2g} = \frac{v^2 \sin 2\theta}{g} \] The $v^2$ and $g$ are common on both sides, so they cancel out.
Step 5: Use the double angle rule.
We know $\sin 2\theta = 2\sin\theta\cos\theta$. Putting this in: \[ \frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta \] The angle is not zero, so $\sin\theta$ is not zero. We can divide both sides by $\sin\theta$. \[ \frac{\sin\theta}{2} = 2\cos\theta \]
Step 6: Solve for $\tan\theta$.
Move terms so that $\sin\theta$ over $\cos\theta$ stays together. \[ \frac{\sin\theta}{\cos\theta} = 4 \] Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, we get the answer. \[ \boxed{\tan\theta = 4} \]