Question

Ratio of the ranges of the bullets fired from a gun at angle θ, 2θ and 4θ is found in the ratio x : 2 : 2, then the value of x will be (Assume same speed of bullets)
A) 1
B) 2
C)√3
D) none of these.

Answer 1

The ranges of the bullets fired from a gun at angle $\theta$, $2\theta$ and $4\theta$ are found in the ratio $x:2:2$, then the value of $x$ will be (Assume same speed of bullets)

A) 1

B) 2

C) $\sqrt{3}$

D) none of these.

The total horizontal distance travelled by the projectile equals:

$$d = \frac{v^2 \sin(2\alpha)}{g}$$

g - the gravitational acceleration

$\alpha$ - the angle at which the projectile is launched

$\nu$ - the velocity at which the projectile is launched

If ratio of the ranges of the bullets fired from a gun at $2\theta$ and $4\theta$ is found in the ratio $2:2$, then:

$$\sin 4\theta = \sin 8\theta \Rightarrow \theta = 15 \quad (\sin 6\theta = \sin 12\theta = \frac{\sqrt{3}}{2})$$

Range of the bullet fired from a gun at angle $\theta$ equals:

$$d = \frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin(3\theta)}{g} = \frac{v^2}{2g}$$

Range of the bullet fired from a gun at angle $2\theta$ equals:

$$d = \frac{v^2 \sin(4\theta)}{g} = \frac{v^2 \sin(6\theta)}{g} = \frac{\sqrt{3}v^2}{2g}$$

Ratio of the ranges equals:

1: $\sqrt{3}$ or $\frac{2}{\sqrt{3}}:2$

Answer: D) none of these.