A particle is projected from ground with speed 80 m/s at an angle `30^(@)` with horizontal from ground. The magnitude of average velocity of particle in time interval t = 2 s to t = 6 s is [Take g = 10 `m//s^(2)] `

A particle is projected from the ground with an initial velocity of 20m//s at an angle of 30^(@) with horizontal. The magnitude of change in velocity in a time interval from t=0 to t=0.5s ( g=10m//s^(2) )

A particle is projected from the surface of the earth with a speed of 20 m* s^(-1) at an angle 30^@ with the horizontal . The range of that particle is

A particle is projected from ground at an angle 45^(@) with initial velocity 20sqrt2 ms^(-1) .The magnitude of average velocity in a timer interval from t=0 to t=3 s in ms^(-1) is

A particle is projected from the surface of the earth with a speed of 20 m* s^(-1) at an angle 30^@ with the horizontal . The time of flight of that particle is

A particle is project from ground with a speed of 6 m//s at an angle of cos^(-1)(sqrt((7)/(27))) with horizontal. Then the magnitude of average velocity of particle from its starting point to the highest point of its trajectory in m//s is:

A particle is projected from ground with velocity 40sqrt(2)m//s at 45^(@) . At time t=2s

A particle is projected from ground with velocity 40sqrt(2)m//s at 45^(@) . At time t=2s

A particle is projected from ground with velocity 40sqrt2m at 45^(@) .At time t=2s

A particle is projected from ground at angle 45^@ with initial velocity 20(sqrt2) m//s . Find (a)change in velocity, (b)magnitude of average velocity in a time interval from t=0 to t=3s .