Galileo writes that for angles of projection of a projectile at angles (45 + θ ) and (45 - θ ) , h

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3-2.Motion in Plane

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

A

$2:1$

B

$1:2$

C

$1:1$

D

$2:3$

Solution

(c) For angle $(45^\circ – \theta )$, $R = \frac{{{u^2}\sin (90^\circ – 2\theta )}}{g} = \frac{{{u^2}\cos 2\theta }}{g}$

For angle $(45^\circ + \theta )$, $R = \frac{{{u^2}\sin (90^\circ + 2\theta )}}{g} = \frac{{{u^2}\cos 2\theta }}{g}$

Standard 11

Physics

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