 # 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?

9 years ago

final angle of velocity vector is 30 with horizontal

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

9 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

9 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.

2 years 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