two particles move near surface of earth with uniform accelerationg = 10m/s2 towards the ground. At initial moment, the particles were located at one point in space and moved with velocities v1 = 3m/s and v2 = 4m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

simply do thier dot product 0 and find the ans..

Consider one particle is moving in direction of +ve x axis with velocity v1 and other in direction of -ve x axis with velocity v2. At time t,velocities of two bodies are: v1=(3)i - (10t)j v2=(-4)i - (10t)j When velocity vectors are perpendicular ,v1.v2=0 So -12+100t2=0 or t=√3/5

Consider one particle is moving in direction of +ve x axis with velocity v₁ and other in direction of -ve x axis with velocity v₂.

At time t,velocities of two bodies are:

v₁=(3)i - (10t)j

v₂=(-4)i - (10t)j

When velocity vectors are perpendicular ,v₁.v₂=0

So

-12+100t²=0

or t=√3/5

At t=√3/5, the bodies would have fallen the same distance or in other words the bodies will always be in a horizontal line...

So the distance between two bodies is the distance between their x co-ordinates which are:

3√3/5 and -4√3/5 respectively..

So the distance is