Question

Galileo writes that for angle of projection of a projectile at angle ( $45^{0} - θ$ ) and ( $45^{0} + θ$ ), the horizontal ranges described by the projectile are in the ratio of (if $θ \leq 45^{0}$ )

Moderate

Solution

For angle ( $45^{0} - θ$ ), $R = \frac{u^{2} \sin (90 ° - 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$
For angle ( $45^{0} + θ$ ), $R = \frac{u^{2} \sin (90 ° + 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$
Thus, ratio becomes 1 : 1.

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