1. Let the two projection angles be $\theta_1$ and $\theta_2$.

   Since both particles are projected with the same speed $u$ and have the same range $R$, we use $R=\frac{u^2\sin 2\theta}{g}.$

   Thus, $\frac{u^2\sin 2\theta_1}{g}=\frac{u^2\sin 2\theta_2}{g}$ which gives $\sin 2\theta_1=\sin 2\theta_2.$

2. For different angles with same range, the projectile angles are complementary: $\theta_2=90^\circ-\theta_1.$

3. Maximum height formula: $h=\frac{u^2\sin^2\theta}{2g}.$

   Therefore, $h_1=\frac{u^2\sin^2\theta_1}{2g}, \qquad h_2=\frac{u^2\sin^2\theta_2}{2g}.$

   Since $\theta_2=90^\circ-\theta_1$, $\sin^2\theta_2=\cos^2\theta_1.$

   Hence, $h_1h_2=\left(\frac{u^2}{2g}\right)^2 \sin^2\theta_1\cos^2\theta_1.$

4. Use $\sin^2\theta_1\cos^2\theta_1=\frac{1}{4}\sin^2 2\theta_1.$

   So,

   =\frac{u^4\sin^2 2\theta_1}{16g^2}.$$

5. Now square the range: $R=\frac{u^2\sin 2\theta_1}{g}$ so $R^2=\frac{u^4\sin^2 2\theta_1}{g^2}.$

6. Compare $R^2$ and $h_1h_2$: $h_1h_2=\frac{R^2}{16}$ therefore, $R^2=16h_1h_2.$

7. Checking options:

   • A: $R^2=h_1h_2$ ❌
   • B: $R^2=16h_1h_2$ ✅
   • C: $R^2=4h_1h_2$ ❌
   • D: $R^2=2h_1h_2$ ❌

Therefore, the correct option is B.