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
$R_2=\frac{u^2\sin 2\left(45^{\circ }-\alpha \right)}{g}$
$=\frac{u^2\sin \left(90^{\circ }-2\alpha \right)}{g}$
$=\frac{u^2\cos 2\alpha }{g}$
$\therefore \frac{R_1}{R_2}=1$