projectile motion - Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.

Sandra (last edited 2 years ago) (Student, UG1 Grade)

If a projectile is released at an angle of \theta_0 with a velocity of v_0 ,

Then, range of a projectile R = \frac{v_0^2 \sin 2\theta_0}{g}

(Elevation)Angle exceed \theta_0 = (45 + \alpha)

Angle fall \theta_0 = (45 - \alpha)

Range of a projectile angle exceed:

R=\frac{v_0^2\sin2\left(45+\alpha\right)}{g}

R=\frac{v_0^2\ \sin\left(90+2\alpha\right)}{g}

R = \frac{v_0^2 \cos 2 \alpha}{g} .........................(1)

Range of a projectile angle fall

R'=\frac{v_0^2\sin2\left(45-\alpha\right)}{g}

R'=\frac{v_0^2\ \sin\left(90-2\alpha\right)}{g}

R'=\frac{v_0^2\ \cos2\alpha}{g} .........................(2)

From equation (1) and (2)

R = R'

Hence, ranges are equal for elevations that exceed or fall short of 45 degrees by the same number \alpha .