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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?

6 years ago

Answers : (2)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

dear Rajan


see the pic for solution


All the best.


AKASH GOYAL


AskiitiansExpert-IITD


 


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1584_23499_Photo-0091.jpg

6 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...


V1.V2=0


 t^2 = 3/25 units


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

6 years ago
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