A particle is projected with a velocity `30ms^(-1)` at an angle of `60^(0)` with the horizontal.The vertical component of velocity at its maximum height is

A body is projected with velocity 24 ms^(-1) making an angle 30° with the horizontal. The vertical component of its velocity after 2s is (g=10 ms^(-1) )

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 body is projected with an initial velocity of 58.8 m/s at angle 60° with the vertical. The vertical component of velocity after 2 sec is

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

A body is thrown into air with a velocity 5 m/s making an angle 30^(@) with the horizontal .If the vertical component of the velocity is 5 m/s what is the velocity of the body ? Also find the horizontal component .

An object is projected with a velocity of 30 ms^(-1) at an angle of 60^(@) with the horizontal. Determine the horizontal range covered by the object.

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 with velocity 50 m/s at an angle 60^(@) with the horizontal from the ground. The time after which its velocity will make an angle 45^(@) with the horizontal is

A ball is projected with a velocity 20 sqrt(3) ms^(-1) at angle 60^(@) to the horizontal. The time interval after which the velocity vector will make an angle 30^(@) to the horizontal is (Take, g = 10 ms^(-2))

A particle is projected with a velocity 10 m//s at an angle 37^(@) to the horizontal. Find the location at which the particle is at a height 1 m from point of projection.