Question

prove that a gun will shoot three times as high when its angle is 60 as when it is 30 but the bullet will carry the same horizontal distance.

Answer 1

The horizontal distance is

$$R=\frac{v^2}{g}\sin(2\alpha)$$

$$R_1=\frac{v^2}{g}\sin60=\frac{\sqrt{3}v^2}{2g}$$

$$R_2=\frac{v^2}{g}\sin120=\frac{\sqrt{3}v^2}{2g}$$

Thus,

$$R_1=R_2$$

$$h_1=\frac{v^2}{2g}\sin^230=\frac{v^2}{8g}$$

$$h_2=\frac{v^2}{2g}\sin^260=\frac{3v^2}{8g}=3h_1$$