1. For a projectile projected with speed $u$ at angle $\theta$:

   • Horizontal range, $R = \frac{u^2 \sin 2\theta}{g}$

   • Maximum height, $H = \frac{u^2 \sin^2\theta}{2g}$

2. Given that the horizontal range and maximum height are equal: $R = H$

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

3. Cancel common factors $\frac{u^2}{g}$: $\sin 2\theta = \frac{\sin^2\theta}{2}$

4. Use the identity: $\sin 2\theta = 2\sin\theta\cos\theta$

   Hence, $2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$

5. Multiply by $2$: $4\sin\theta\cos\theta = \sin^2\theta$

6. For non-zero projection angle, divide by $\sin\theta$: $4\cos\theta = \sin\theta$

   Therefore, $\tan\theta = 4$

7. Thus, $\theta = \tan^{-1}(4)$

8. Checking options:

   • A: $\tan^{-1}(1/2)$ ❌
   • B: $\tan^{-1}(2)$ ❌
   • C: $\tan^{-1}(1/4)$ ❌
   • D: $\tan^{-1}(4)$ ✅

Therefore, the correct answer is Option D.