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

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

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

The equation of the trajectory of a projectile on a vertical plane is y= ax-bx^2 , where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and the angle of projection with respect to the horizontal.

For angles theta and (90^@-theta) of projection, a projectile has the same horizontal range R. The maximum heights attained are H_1 and H_2 respectively. Then the relation among R,H_1 and H_2 is

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 ?

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

Equations of trajectory of a projectile are x=4 t, y= 3t - 5 t ^(2) , where x and y are in metre and t is in second . Find (i) maximum height attained by the projectille , (ii) horizontal range of the projectile .

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^@ .