A body is projected obliquely at `30^(0)` with the horizontal .The time when angular momentum about point of projection is maximum in sec is --------- if (u=10m/s `g=10m/s^(2)`

A body is projected with a speed u at an angle of 30^(@) with the horizontal at t=0 . The angular momentum of the body about the point of projection is (xsqrt(3)mu^(3))/(64g) at t=(3u)/(4g) . Find the value of x ?

A body of mass m is projected with a velocity u at an angle theta with the horizontal. The angular momentum of the body, about the point of projection, when it at highest point on its trajectory is

A particle of mass "m" is projected with a velocity ' "u" ' making an angle of 30^(@) with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height "h" is :

A particle of mass 1kg has been thrown with initial speed 20 m/s making an angle 60° with the horizontal ground. the angular momentum of the particle about point of projection when the projectile is at highest point is (g=10m/s^2)

A particle of mass m is thrown with speed v_(0) at angle 30^(@) with horizontal. Find the angular momentum about the point of projection when the particle is at the highest point of its trajectory.

When a body is projected at an angle with the horizontal in the uniform gravitational field of the earth, the angular momentum of the body about the point of projection, as it proceeds along its path

A particle of mass m is projected with a speed u at an angle theta with the horizontal. Find angular momentum of particle after time t about point of projection.

A body of mass m is projected with intial speed u at an angle theta with the horizontal. The change in momentum of body after time t is :-