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a particle is projected at an initial velocity u with a angle of 60 with the vertical. The time after which the velocity vector becomes perpendicular to the initial velocity vector?

a particle is projected at an initial velocity u with a angle of 60 with the vertical. The time after which the velocity vector becomes perpendicular to the initial velocity vector?

Grade:12

4 Answers

himesh kothari
14 Points
8 years ago

final angle of velocity vector is 30 with horizontal

show their horizontal components of velocity same and by vertical motion calc time

chukkaiwar rohit
14 Points
8 years ago

When a projectile makes 600 it makes an angle 300 with the horizontal.the velocity vector of the partical initially is v0=ucos30i+usin30j, after a time t the velocity vector is v=ucos30i+(usin30-gt)j the two vectors are perpendicular so their dot product is zero. we get that ucos30xucos30+usin30xusin30=gtsin30u=>u2=gtsin30=>t=u/gsin30

SOURAV MISHRA
37 Points
8 years ago

(ucos30°i + (usin30° - gt)j).(ucos30°i + usin30°j) = 0

u2 = ugtsin30°

t = 2u/g.

this is the time after which the velocity of the particle becomes perpendicular to the initial velocity vector.

Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

When a projectile makes 600 it makes an angle 300 with the horizontal.the velocity vector of the partical initially is v0=ucos30i+usin30j, after a time t the velocity vector is v=ucos30i+(usin30-gt)j the two vectors are perpendicular so their dot product is zero. we get that ucos30xucos30+usin30xusin30=gtsin30u=>u2=gtsin30=>t=u/gsin30

Thanks and Regards

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