Question

Galileo writes that for angles of projection of a projectile at angles $\left(45^{\circ}+\theta\right)$ and $\left(45^{\circ}-\theta\right)$, the horizontal ranges described by the projectile are in the ratio

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $2:3$

Answer 1

Answer: (C)

$1:1$

Answer 2

Horizontal range $R=\frac{u^{2} \sin 2 \theta}{g}$ $\Rightarrow R \propto \sin 2 \theta$ $\therefore R_{1} \propto \sin 2\left(45^{\circ}+\theta\right)$ $R_{1} \propto \sin \left(90^{\circ}+2 \theta\right)$ Similarly, $R_{2} \propto \sin \left(90^{\circ}-2 \theta\right)$ $\because \sin \left(90^{\circ}+2 \theta\right)=\cos 2 \theta$ $\because \sin \left(90^{\circ}-2 \theta\right)=\cos 2 \theta$ so, $\frac{R_{1}}{R_{2}}=\frac{1}{1}$ or $R_{1}: R_{2}=1: 1$