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Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of $45^o$ by equal amounts, the ranges are equal”. Prove this statement.
Solution
Answer For a projectile launched with velocity
$v _{ o }$ at an angle $\theta_{ o },$ the range is given by
$R=\frac{v_{o}^{2} \sin 2 \theta_{0}}{g}$
Now, for angles, $\left(45^{\circ}+\alpha\right)$ and $\left(45^{\circ}-\alpha\right), 2 \theta_{0}$ is $\left(90^{\circ}+2 \alpha\right)$ and $\left(90^{\circ}-2 \alpha\right),$ respectively. The values of $\sin \left(90^{\circ}+2 \alpha\right)$ and $\sin \left(90^{\circ}-2 \alpha\right)$ are
the same, equal to that of $\cos 2 \alpha .$ Therefore, ranges are equal for elevations which exceed or fall short of $45^{\circ}$ by equal amounts $\alpha$
