JEE Main 2026 — Motion In Two Dimensions Question with Solution

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:
$t_1=\frac{2u\sin \theta }{g}$
$t_2=\frac{2u\cos \theta }{g}$

Multiplying $t_1$ and $t_2$, we get:
$t_1{t}_2=\left(\frac{2u\sin \theta }{g}\right)\left(\frac{2u\cos \theta }{g}\right)=\frac{2}{g}\left(\frac{u^2(2\sin \theta \cos \theta )}{g}\right)$

Since the horizontal range is $R=\frac{u^2\sin 2\theta }{g}$, we can write:
$t_1{t}_2=\frac{2R}{g}$

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

$50=\frac{2R}{10}$

$2R=500\Rightarrow R=250$ m

Answer: $250$