A particle is projected with speed `10 m//s` at angle `60^(@)` with the horizontal. Then the time after which its speed becomes half of initial.

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 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 particle is projected with speed 100m/s at angle theta=60^(@) with the horizontal at time t=0" . At time 't' the velocity vector of the particle becomes perpendicular to the direction of velocity of projection.Its tangential acceleration at time 't' is (g =10m/s^(2))

A particle is projected with a speed of "18m/s" at an angle of 60^(@) with the horizontal.The average velocity of the particle between two instants when it is at the same height is

A particle is projected with speed 'u' at an angle 60^@ with horizontal, then find the speed when its angle with horizontal changes to 30^@ .

A particle is projected with a speed of 10 m//s at an angle 37^(@) from horizontal, angular speed to particle with respect to point of projection at t = 0.5 sec . Is ((16X)/(61)) rad/sec then calculate X :

A particle is projected with a speed of 10 m//s at an angle 37^(@) from horizontal, anglular speed to particle with respect to point of projection at t = 0.5 sec . Is ((16X)/(61)) rad/sec then calculate X :

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) .