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rajan jha Grade: 12

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?

7 years ago

Answers : (2)

AKASH GOYAL AskiitiansExpert-IITD
419 Points

dear Rajan

see the pic for solution

All the best.




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7 years ago
vikas askiitian expert
510 Points

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...


 t^2 = 3/25 units

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

7 years ago
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