Question

A particle is projected with velocity 10 metre per sec at an angle of 60 degree from ground. then the vertical component of velocity vector when instantaneous velocity vector becomes perpendicular to initial velocity​

Answer 1

Answer:

-5√3

Explanation:

A particle is projected with velocity 10 metre per sec at an angle of 60 degree from ground. then the vertical component of velocity vector when instantaneous velocity vector becomes perpendicular to initial velocity​

Tanα = VSinα/VCosα

Tanα = Tan 60° = √3

VSinα = Vertical Velocity = 10√3/2 = 5√3 m/s

VCosα = Horizontal Velocity = V/2 = 5 m/s

Horizontal Velocity will remain constant = 5 m/s

instantaneous velocity vector becomes perpendicular to initial velocity​

=> β = α ± 90° = 60  ± 90° = -30° or 150°

=> Vertical Velocity = VSin(-30°) = -V/2

Horizontal Velocity = VCos(-30°) = V√3/2

V√3/2 = 5

=>  V =10/√3

=> -V/2 = -5/√3

hence β = 150°

Vertical Velocity = VSin150° = V/2

Horizontal Velocity = VCos150° = -V√3/2

-V√3/2 = 5

=> V= -10/√3

=>  V/2 = -5/√3

vertical component of velocity = -5√3

Answer 2

Explanation:

therefore

V/2 = -5/ROOT3