A ball is projected with `20sqrt2 m//s` at angle `45^(@)` with horizontal.The angular velocity of the particle at highest point of its journey about point of projection is

A particle of mass m is projected with a velocity mu at an angle of theta with horizontal.The angular momentum of the particle about the highest point of its trajectory is equal to :

A particle of mass m is projected with a velocity v at an angle of theta with horizontal. The angular momentum of the particle at the highest point of its trajectory is equal to :

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

A particle of mass m=1kg is projected with speed u=20sqrt(2)m//s at angle theta=45^(@) with horizontal find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

A particle of mass m=1kg is projected with speed u=20sqrt(2)m//s at angle theta=45^(@) with horizontal find the torque of the weight of the particle about the point of projection when the particle is at the highest point.