Question

Two particles move in a uniform gravitational field with an acceleration $g$. At the initial moment the particles were located over a tower at one point and moved with velocities $v_1=3m/s$ and $v_2=4m/s$ horizontally in opposite directions. The distance between the particles at the moment when their velocity vectors become mutually perpendicular, rounded off up to one decimal place will be: ( Take $g=9.8m{s}^{-2}$)

• A. 2.5 m
• B. 5 m
• C. 8 m
• D. None of these

Answer 1

The correct option is A 2.5 m

Let the particles project as given in above figure.

The vertical velocities of both the particles after time $t$ would be

$v=0+gt=gt$

Let $i,j$ the angles the projectiles form with vertical axis.

$⟹\tan i=\frac{-3}{gt}$ and $\tan j=\frac{4}{gt}$.

When the projectiles are perpendicular to each other, $i+j={90}^{\circ }$

$\tan (i+j)=\frac{\tan i+\tan j}{1+\tan i\tan j}=\infty$

$⟹\tan i\tan j=-1$

$⟹gt=\sqrt{12}$

Distance between the particles = $3t+4t=7t=7\frac{\sqrt{12}}{g}=2.5m$