Let  ‘ϴ’ be the angle of elevation. When ϴ=60o, then maximum height,\\[H_1 = \\frac{u^2 sin^2 60^0}{2g} = \\frac{3u^2}{8g} \\]where ‘u’ is the initial velocity of the bullet, ‘g’ is the gravitational force of attraction.  Horizontal Range,\\[R_1 = \\frac{u^2 sin(2 x 60^0)}{g} = \\frac{\\sqrt{3u^2}}{2g} \\]When ϴ=30o, then maximum height,\\[H_2 = \\frac{u^2 sin^2 30^0}{2g} = \\frac{u^2}{8g} \\]Horizontal Range, \\[R_2 = \\frac{u^2 sin(2 x 30^0)}{g} = \\frac{\\sqrt{3u^2}}{2g} \\]Hence, \\[H_1 = 3H_2 \\] and \\[R_1 = R_2 \\]   Therefore a  gun will shoot three times as high when its angle of elevation is as when it is but covers the same horizontal range.