A particle is projected with a velocity `vecv=8hati+6hatj m//s`. The time after which it will starts moving perpendicular to its initial direction of motions is

A particle is projected from ground with initial velocity 3hati + 4hatj m/s. Find range of the projectile :-

A projectile is projected with initial velocity (6hati+8hatj) m/s.Minimum time after which its velocity becomes parallel to the line 4y-3x+2=0, is ?( bar(g)=-10hatj m/ s^(2) )

A particle is projected upwards with a velocity of 110m/sec at an angle of 60^(@) with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g =10m//sec^(2)

A particle is projected upwards with a velocity of 100 m//sec at an angle of 60^(@) with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10m//sec^(2) .

A particle is projected upwards with a velocity of 100 m//s at an angle of 60^(@) with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m//s^(2) .

A particle is projected upwards with a velocity of 100 m//s at an angle of 60^(@) with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m//s^(2) .

A vector is given by vecV=(8.0hati+2.0hatj)m//s . What will be the unit vector perpendicular to vecV ?

A particle is projected with a velocity of 20sqrt(3) ms^(-1) at an angle of 60^(@) to the horizontal. At the moment, the direction of motion of the projectile is perpendicular to the direction of initial velocity of projection, its velocity is

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))