Question

If the angles of projection of a projectile with same initial velocity exceed or fall short of $45^{\circ}$ by equal amount $\alpha$, then the ratio of horizontal ranges is

• A. 1: 2
• B. $1: \sqrt{2}$
• C. 1: 4
• D. 1: 1

Answer 1

Answer: (D)

Let u be the initial velocity of the projectile.
For angle of projection $\left(45^{\circ}+\alpha\right)$, horizontal range is
$R_{1}=\frac{u^{2} \sin 2\left(45^{\circ}+\alpha\right)}{g}$
$=\frac{u^{2} \sin \left(90^{\circ}+2 \alpha\right)}{g}$
$=\frac{u^{2} \cos 2 \alpha}{g}$
For angle of projection $\left(45^{\circ}-\alpha\right)$, horizontal range is
$R_{2} =\frac{u^{2} \sin 2\left(45^{\circ}-\alpha\right)}{g}$
$=\frac{u^{2} \sin \left(90^{\circ}-2 \alpha\right)}{g}$
$=\frac{u^{2} \cos 2 \alpha}{g}$
$\therefore \frac{R_{1}}{R_{2}} =1$