A particle is projected with initial velocity of `hati+2hatj`. The eqaution of trajectory is `(take g=10ms^(-2))`

A projectile is given an initial velocity hati+2 hatj .The cartesin equation of its path is (g=10ms^(-2))

A projectile is given an initial velocity hati+2 hatj .The cartesin equation of its path is (g=10ms^(-2))

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

A particle is projected from ground with initial velocity 3hati + 4hatj m/s. Find range of the projectile :-

A body is projected with a velocity vecv =(3hati +4hatj) ms^(-1) The maximum height attained by the body is: (g=10 ms^(-2))

A body is projected with a velocity vecv =(3hati +4hatj) ms^(-1) The maximum height attained by the body is: (g=10 ms^(-2))

What are the angles of projection of a projectile projected with velocity 30 ms^(-1) , so that the horizontal range is 45 m ? Take g = 10 ms^(-2) .

A body is projected vertically upward with the velocity v=3hati+4hatjms^(-1) . The maximum height attained by the body is ( g=10ms^(-2) ).