Show that a gun will shoot woll shoot three times as hith when elevated at an angle of ` 60^(@)` as when fired at an angle ` 30^(@)` but will have the same horizontal rande.

Explanation

Let (u) be the velocity of shot fired from a gun at an angle ` theta` with the horizontal. Then
Max. height , `H= (u^2 sin^@ theta)/ (2 g) ` ,
Horizontal range, ` R = 9 u^2 sin 2 theta) /g`
When ` theta=30^@` ,
` H_1 = (u^2 sin ^2 30^@) /(2 g) = (u^2 sin xx 1//4) /(2 g) = u^2/( 8g)`
` R_1 =(u^2 sin 2 xx 30^@)/ ( g) = (u^2 sin 60^@)/g = (u^2 sqrt 3)/ (g 2)`
Whem `theta = 60^@`,
H_2 = (u^2 sin ^2 60^@)/(2 g) = (u^2 (sqrt 3//2)^2)/(2 g) = (u^2 sqrt 3)/ (g 2)`
` R_2 = (u^2 sin 2 xx 60^@_ /g =( u^2 sin 120^@_ g = (3 u^@)/( 8 2)`
From above, we bote that ` R_1 = R_2` and
` H_2 /H_1 = (23 u^2 //8 g)/(u^2 // 8 g/(u^1 // 8 g) = 3 or H_2 =3 H_1 (Proved)`.