let us solve this question using vectors.
let the first particle be thrown horizontally in +x direction, so u1=4 i m/s
and for the other particle so velocity is u2= -9 i m/s
let us find their velocity vectors after t seconds, by using the formula v=u+a*t.
as we know that the body’s horizontal velocity will not be changed because no force acts on horizontal only vertical velocity will be changed.
so vertical velocity gained by first body u=0, a=-10m/s2, v1= -10t m/s
and v2 also equal to -10t m/s
so the velocity vector of v1 and v2 after t seconds is:
V1= 4i – 10t j, V2= -9i -10t j;
we know that the dot product of 2 vectors is 0 when they are perpendicular to each other.
so here the velocity vectors are perpendicular, so their dot product should also be 0
by doing dot product and equating it with 0 we get, t=0.6 sec,
so the displacement vectors at time 0.6 sec will be,
for the first particle:
Sx=4*0.6=2.4
Sy=-0.5*a*t2=-1.8 m
so S1=2.4i-1.8j
similarly we can find S2 and it comes out to be:
S2= -5.4i-1.8j
so the separation between them at the time is 2.4+5.4=6.8 m
(taking only x direction displacement as the y direction displacement is same.
so the answer is 6.8m
thanks
