If the horizontal range is 4 times of the maximum height then what is the angle of projection?

If the range of a projectile, projected from the surface of the earth is four times the maximum height attained, what is the angle of projection ?

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

The resultant velocity of a projectile at its highest point is sqrt(6/7) times that at half the maximum height. Show that the angle of projection with respect to the horizontal is 30^@ .

A stone is thrown up from the earth's surface with an initial velocity of 28 m*s^(-1) , making and angle 30^@ with the horizontal. Find the maximum height attained and the range of the projectile motion of the stone.

Statement I: The initial velocity of projectile =(ahati+bhatj) . The horizontal range becomes maximum for a=b. Statement II: for the same magnitude of initial velocity , the horizontal range of a projectile becomes maximum for the angle 45^@ of projection.

If the length of the shadow of a vertical pole on the horizontal grounds is sqrt3 times its height, then the angle of elevation of the sun is

For a projectile, ("horizontal range")^2 =48xx("maximum height")^2 The angle of projection is

Statement I: The horizontal range of a projectile is the same for angles 30 ^@ and 60^@ of projection. Statement II: The horizontal range of projectile is independent of the angle of projection.