Question

Two particles move near surface of earth with uniform acceleration g=10m/s2 towards the ground. At the initial moment, the particles were located at one point in space and moved with velocities V1=3.0m/s and V2= 4.0m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Answer 1

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..

Hope it helps

Answer 2

Answer:

hope it gelps.the answer is bit incorrect tho