# two particles move near surface of earth with uniform acceleration g = 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.

Prakhar Srivastava
18 Points
11 years ago

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

mahendra shah
19 Points
11 years ago

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

sonal sharma
31 Points
11 years ago

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