Two guns A and B can fire bullets at speeds $v_A = 1$ km/s and $v_B = 2$ km/s, respectively. Both guns are fired from a height $H$ above a horizontal ground, where $H = \frac{v_A^2}{2g}$. They are fired in all possible directions. The ratio of maximum areas covered by the bullets on the ground fired by the two guns is
1 : 10
Maximum Area Covered by Bullets from Guns A and B
This solution explains how to find the ratio of the maximum areas covered by bullets fired from two guns, A and B, launched from the same height but with different initial speeds. We will analyze the physics of projectile motion to determine the range and then the area.
Understanding Projectile Motion from Height
When a projectile is fired from a height $H$ above the ground with an initial speed $v$, its trajectory depends on the launch angle. The phrase "fired in all possible directions" implies we need to consider the angle that results in the maximum horizontal range ($R_{max}$). The area covered on the ground is related to this maximum range, typically calculated as $A = \pi R_{max}^2$, assuming the area is circular with radius $R_{max}$.
Given Information
We are provided with the following details:
- Initial speed of gun A: $v_A = 1$ km/s
- Initial speed of gun B: $v_B = 2$ km/s
- Firing height: $H$, related to $v_A$ by the equation $H = \frac{v_A^2}{2g}$
- $g$ represents the acceleration due to gravity.
Calculating the Height $H$
The height $H$ is defined in terms of $v_A$ and $g$: $H = \frac{v_A^2}{2g}$ Multiplying both sides by $2g$, we get a useful relationship: $2gH = v_A^2$ This relationship will simplify our calculations for the maximum range.
Formula for Maximum Horizontal Range ($R_{max}$)
For a projectile launched from a height $H$ with an initial speed $v$, the maximum horizontal range $R_{max}$ achieved when considering all possible launch directions is given by the formula:
$R_{max} = \frac{v}{g} \sqrt{v^2 + 2gH}$This formula represents the furthest horizontal distance the projectile can travel before hitting the ground. The term $\sqrt{v^2 + 2gH}$ relates to the final speed of the projectile upon impact, derived from energy conservation.
Calculating Maximum Range for Gun A ($R_A$)
We apply the $R_{max}$ formula using the speed $v_A$ for gun A. We also substitute $2gH = v_A^2$: $R_A = \frac{v_A}{g} \sqrt{v_A^2 + 2gH}$ Substitute $2gH = v_A^2$: $R_A = \frac{v_A}{g} \sqrt{v_A^2 + v_A^2}$ $R_A = \frac{v_A}{g} \sqrt{2v_A^2}$ Simplify the square root: $R_A = \frac{v_A}{g} (\sqrt{2} v_A)$ $R_A = \frac{\sqrt{2} v_A^2}{g}$
Calculating Maximum Range for Gun B ($R_B$)
Similarly, we apply the $R_{max}$ formula for gun B, using its speed $v_B = 2v_A$, and the same height $H$ (so $2gH = v_A^2$ still holds): $R_B = \frac{v_B}{g} \sqrt{v_B^2 + 2gH}$ Substitute $v_B = 2v_A$ and $2gH = v_A^2$: $R_B = \frac{2v_A}{g} \sqrt{(2v_A)^2 + v_A^2}$ $R_B = \frac{2v_A}{g} \sqrt{4v_A^2 + v_A^2}$ $R_B = \frac{2v_A}{g} \sqrt{5v_A^2}$ Simplify the square root: $R_B = \frac{2v_A}{g} (\sqrt{5} v_A)$ $R_B = \frac{2\sqrt{5} v_A^2}{g}$
Ratio of Maximum Ranges
To find the ratio of the areas, we first find the ratio of the maximum ranges ($R_A$ to $R_B$):
$\frac{R_A}{R_B} = \frac{\frac{\sqrt{2} v_A^2}{g}}{\frac{2\sqrt{5} v_A^2}{g}}$Cancel out the common terms $\frac{v_A^2}{g}$:
$\frac{R_A}{R_B} = \frac{\sqrt{2}}{2\sqrt{5}}$Simplify the expression:
$\frac{R_A}{R_B} = \frac{\sqrt{2}}{(\sqrt{2}\sqrt{2})\sqrt{5}} = \frac{1}{\sqrt{2}\sqrt{5}} = \frac{1}{\sqrt{10}}$Ratio of Maximum Areas Covered
The maximum area covered by each gun ($A_A$ and $A_B$) is given by $A = \pi R_{max}^2$. The ratio of the areas is:
$\frac{A_A}{A_B} = \frac{\pi R_A^2}{\pi R_B^2}$This simplifies to the square of the ratio of the ranges:
$\frac{A_A}{A_B} = \left(\frac{R_A}{R_B}\right)^2$Substitute the ratio of the ranges we calculated:
$\frac{A_A}{A_B} = \left(\frac{1}{\sqrt{10}}\right)^2$ $\frac{A_A}{A_B} = \frac{1}{10}$Conclusion
The ratio of the maximum areas covered by the bullets fired from gun A compared to gun B is 1:10.
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