Horizontal Range Given Complementary Projection Times of Flight

To find the horizontal range $R$ of the projectiles, we utilize the relationship between the range and the times of flight for complementary angles of projection.

For two bodies projected with the same initial speed $u$ to have the same horizontal range $R$, their angles of projection must be complementary, i.e., $\theta$ and $(90^\circ - \theta)$.

1. The expressions for the times of flight for the two projectiles are given by: $T_1 = \frac{2u \sin\theta}{g}$ $T_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos\theta}{g}$

2. The expression for the horizontal range $R$ is: $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$

3. Multiplying the two times of flight $T_1$ and $T_2$: $T_1 T_2 = \left(\frac{2u \sin\theta}{g}\right) \left(\frac{2u \cos\theta}{g}\right) = \frac{2}{g} \left(\frac{2u^2 \sin\theta \cos\theta}{g}\right) = \frac{2R}{g}$

4. Rearranging the equation to solve for $R$: $R = \frac{1}{2} g T_1 T_2$

5. Substituting the given values $T_1 = 5\text{ s}$, $T_2 = 10\text{ s}$, and $g = 10\text{ m/s}^2$: $R = \frac{1}{2} \times 10 \times 5 \times 10 = 250\text{ m}$

Thus, the value of $R$ is $250\text{ m}$, which corresponds to Option A.