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two particle move in uniform gravitational field with an accelration 'g'.At the initial moment ,the particle were located over a tower at one point and moves with velocity v1=3m/s and v2=4m/s horizontally in opposite direction. find distance between them at the moment when thier velocity vector become mutually perpendicular?

two particle move in uniform gravitational field with an accelration 'g'.At the initial moment ,the particle were located over a tower at one point and moves with velocity v1=3m/s and v2=4m/s horizontally in opposite direction. find distance between them at the moment when thier velocity vector become mutually perpendicular?

Grade:12

2 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
10 years ago

dear Rajan

see the pic for solution

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

 

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vikas askiitian expert
509 Points
10 years ago

at any time t

 velocity of first particle is V1 =3i - gt j

position of first paticle is S1= 3ti -gt^2/2 j

 

 velocity of second particle is V2= -4i -gt j

position of second particle is S2 =-4t -gt^2/2 j                                     (i and j are unit vectors along x axis and y axis)

if their velocities are mutually perpendicular then dot  product will will be 0...

V1.V2=0

 t^2 = 3/25 units

distance bw them at dis time is S1-S2=5t= (7sqrt3)/5 units

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