Two particles are simultaneously projected in the horizontal direction from a point P at a certain height. The initial velocities of the particles are oppositely directed to each other and have magnitude v each. The separation between the particles at a time when their position vectors (drawn from the point P) are mutually perpendicular, is

Question

Two particles are simultaneously projected in the horizontal direction from a point P at a certain height. The initial velocities of the particles are oppositely directed to each other and have magnitude v each. The separation between the particles at a time when their position vectors (drawn from the point P) are mutually perpendicular, is (A) (v2/2g) (B) (v2/g) (C) (4v2/g) (D) (2v2/g)

Tardigrade Admin · Accepted Answer

vecr1 =vt hati -(1/2) gt2 hatj, vecr2 =vt (- hati)-(1/2)gt2 hatj vecr1 r2 =0 , ⇒ -v2t2 +(1/4)g2t4 =0, v2=(1/4)g2t2, v=(gt/2), Δ x =2vt =2× v×(2v/g) =(4v2/g)

$\frac{v ^{2}}{2 g}$

$\frac{v ^{2}}{g}$

$\frac{4 v ^{2}}{g}$

$\frac{2 v ^{2}}{g}$

2gv2​

gv2​

g4v2​

g2v2​

r1​=vti^−21​gt2j^​,r2​=vt(−i^)−21​gt2j^​ r1​​r2​​=0,⇒−v2t2+41​g2t4=0, v2=41​g2t2,v=2gt​, Δx=2vt=2×v×g2v​=g4v2​

$\frac{v ^{2}}{2 g}$

$\frac{v ^{2}}{g}$

$\frac{4 v ^{2}}{g}$

$\frac{2 v ^{2}}{g}$