A particle is projected with a velocity of `20sqrt(3)`m`s^(-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?

Two particle are projected with same initial velocities at an angle 30^(@) and 60^(@) with the horizontal .Then

A particle is projected with speed 100m/s at angle theta=60^(@) with the horizontal at time t=0" . At time 't' the velocity vector of the particle becomes perpendicular to the direction of velocity of projection.Its tangential acceleration at time 't' is (g =10m/s^(2))

A particle is projected with a velocity of 20 ms^(-1) at an angle of 60^(@) to the horizontal. The particle hits the horizontal plane again during its journey. What will be the time of impact ?

A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

A ball is thrown with velocity 8 ms^(-1) making an angle 60^(@) with the horizontal.Its velocity will be perpendicular to the direction of initial velocity of projection after a time of

A particle is projected with velocity 20 ms ^(-1) at angle 60^@ with horizontal . The radius of curvature of trajectory , at the instant when velocity of projectile become perpendicular to velocity of projection is , (g=10 ms ^(-1))

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)