Question

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

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

Answer 1

Answer: (C)

1 : 1

Answer 2

For a projectile launched with velocity u at an angle $\theta,$ the horizontal range is given by

$R = \frac{u^{2}\sin 2\theta}{g}$

(i) For$\theta_{1} = 45{^\circ} - \theta$

$R_{1} = \frac{u^{2}\sin(90{^\circ} - 2\theta)}{g} = \frac{u^{2}\cos 2\theta}{g}$

(ii) For $\theta_{2} = 45^{0} + \theta$

$R_{2} = \frac{u^{2}\sin(90{^\circ} + 2\theta)}{g} = \frac{u^{2}\cos 2\theta}{g} = R_{1} \therefore\frac{R_{2}}{R_{1}} = 1$