A particle of mass 1 kg is projected at an angle of `30^(@)` with horizontal with velocity v = 40 m/s . The change in linear momentum of the particle after time t = 1 s will be (g = 10 `m// s^(2)`)

A particle is projected at an angle 60^(@) with the horizontal with a speed 10m//s . Then, latus rectum is (Take g=10m//s^(2) )

A body is projected at angle 30° to horizontal with a velocity 50 msl. Its time of flight is (g=10m//s^(2))

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 particle is projected from the ground at an angle of 60^(@) with horizontal at speed u = 20 m//s. The radius of curvature of the path of the particle, when its velocity. makes an angle of 30^(@) with horizontal is : (g=10 m//s^(2)

A particle is projected at an angle 60^@ with horizontal with a speed v = 20 m//s . Taking g = 10m//s^2 . Find the time after which the speed of the particle remains half of its initial speed.

A particle is projected at an angle 60^@ with horizontal with a speed v = 20 m//s . Taking g = 10m//s^2 . Find the time after which the speed of the particle remains half of its initial speed.

An object of mass 10 kg is projected from ground with speed 40 m/s at an angle 60^(@) with horizontal the rate of change of momentum of object one second after projection in SI unit is [ Take g = 9.8 m/ s^(2) ]

An object of mass 10 kg is projected from ground with speed 40 m/s at an angle 60^(@) with horizontal the rate of change of momentum of object one second after projection in SI unit is [ Take g = 9.8 m/ s^(2) ]