A particle is projected with a speed `10sqrt(2)` making an angle `45^(@)` with the horizontal . Neglect the effect of air friction. Then after 1 second of projection . Take `g=10m//s^(2)`.

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

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=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.