The speed of a projectile (u) rekuces by ` 50 %` on reachig maximum hight. What is the range on the horizontal plane ?

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

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 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 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 sqrt(3)/(2) times of its initial speed 'u' of projection Its range on the horizontal plane:-

For the angular projessction theta , the velcity of projectile is (u). Let (H) be the maximum heitht reached by the projectile and ® be its horizontal range , show that R^2/(8 H) + 2 H is equal to its maximum range.

The maximum height attaine by a projectile is increased by 10% by increasing its speed of projecton, without changing the angle of projection. What will the percentage increase in the horizontal range.