The speed at the maximum height of a projectile is `sqrt(3)/(2)` times of its initial speed 'u' of projection Its range on the horizontal plane:-

The speed at the maximum height of a projectile is (1)/(3) of its initial speed u. If its range on the horizontal plane is (n sqrt(2)u^(2))/(9g) then value of n is

The speed at the maximum height of a projectile is (1)/(3) of its initial speed u. If its range on the horizontal plane is (n sqrt(2)u^(2))/(9g) then value of n is

The velocity at the maximum height of a projectile is sqrt(3)/2 times the initial velocity of projection. The angle of projection is

The velocity at the maximum height of a projectile is sqrt(3)/2 times the initial velocity of projection. The angle of projection is

The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is

The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is

The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is