[" Show that the maximum height reached by a projectile launched at an angle of "45^(@)" is one "],[" quarter of its range."]

Show that the maximum height attained by a projectile is one fourth of its maximum horizontal range.

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 range of particle when launched at an angle of 15^(@) with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45^(@) to the horizontal

The range of a projectile launched at an angle of 15^(@) with horizontal is 1.5km. The range of projectile when launched at an angle of 45^(@) to the horizontal is

The range of a projectile launched at an angle of 15^(@) with horizontal is 1.5km. The range of projectile when launched at an angle of 45^(@) to the horizontal is

The range of a projectile launched at an angle of 15^(@) with horizontal is 1.5km. The range of projectile when launched at an angle of 45^(@) to the horizontal is

The range of a projectile, when launched at an angle of 15^(@) with the horizontal is 1.5 km. what is the range of the projectile, when launched at an angle of 45^(@) to the horizontal with the same speed ?

The range of a projectile, when launched at an angle of 15^(@) with the horizontal is 1.5 km. what is the range of the projectile, when launched at an angle of 45^(@) to the horizontal with the same speed ?

The range of a particle when launched at an angle of 15^@ with horizontal is 1.5 km.What is the range of the projectile when launched at an angle of 45^@ to the horizonta?