Question

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

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

Answer 1

Answer: (C)

$1:1$

$1)$ For $(45+a)$

Range $=\frac{u^{2}sin\left(2a\right)}{g}$

$Range\left(1\right)=\frac{u^{2}sin(2\times (45+a\left)\right)}{g}$

$=\frac{u^{2}cos2a}{g}$

$2)$ For $(45-a)$

$Range\left(2\right)=\frac{u^{2}sin(90-2a)}{g}$

$=\frac{u^{2}cos2a}{g}$

$\frac{Range\left(1\right)}{Range\left(2\right)}=\frac{1}{1}=1:1$