Two identical bodies, projected with the same speed at two different angles cover the same horizontal range \(R\). If the time of flight of these bodies are \(5\) s and \(10\) s, respectively, then the value of \(R\) is ________ m. (Take \(g=10\) m/s\(^2\))

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

The times of flight for these two angles are given by:
\(t_1 = \dfrac{2u \sin \theta}{g}\)
\(t_2 = \dfrac{2u \cos \theta}{g}\)

Multiplying \(t_1\) and \(t_2\), we get:
\(t_1 t_2 = \left(\dfrac{2u \sin \theta}{g}\right) \left(\dfrac{2u \cos \theta}{g}\right) = \dfrac{2}{g} \left(\dfrac{u^2 (2 \sin \theta \cos \theta)}{g}\right)\)

Since the horizontal range is \(R = \dfrac{u^2 \sin 2\theta}{g}\), we can write:
\(t_1 t_2 = \dfrac{2R}{g}\)

Given \(t_1 = 5\) s, \(t_2 = 10\) s, and \(g = 10\) m/s\(^2\), substituting these values:
\(5 \times 10 = \dfrac{2R}{10}\)

\(50 = \dfrac{2R}{10}\)

\(2R = 500 \Rightarrow R = 250\) m

Answer: \(250\)