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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.
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
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