A particle is projected from a point P(0, 2 ,0) m with a velocity 10m/s making an angle `45^(@)` with the horizontal.The plane of projectile motion passes through a horizontal line PQ which makes an angle of `37^(@)` with the positive x -axis XY plane is horizontal. The coordinates of the point where the particle will strike the line PQ is:

A particle is projected from a point P (2,0,0)m with a velocity 10(m)/(s) making an angle 45^@ with the horizontal. The plane of projectile motion passes through a horizontal line PQ which makes an angle of 37^@ with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle will strike the line PQ is

A particle is projected from a point P(2, 0, 0)m with a velocity 10 m//s making an angle 45^(@) with the horizontal. The plane of projectile motion passes through a horizontal line PQ which an angle of 37^(@) with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle will strike the line PQ is :- ("take "g=10 m//s^(2))

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 particle is projected with velocity 10m/s at an angle of 45^(@) with the horizontal from some height. Consider point of projection as origin and motion in xy plane. Then (vecg=10hatj m//s^(2))

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 projectile is projected from horizontal with velocity (u) making an angle 45^(@) with the horizontal direction. Find the distance of the highest point of the projectile from its starting point.

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.