Two identical bodies, projected with the same speed at two...

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: $250$