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A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of \(H_{\mathrm{A}}\) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \(H_{\mathrm{B}}\) below the center. Find the ratio \(H_{\mathrm{B}} / H_{\mathrm{A}}\).

Short Answer

Expert verified
The ratio \(H_{\mathrm{B}} / H_{\mathrm{A}}\) is 4.

Step by step solution

01

Analyze the Projectile Motion

When the rifle is aimed parallel to the ground, the bullet's horizontal velocity is constant, but it is subject to gravitational acceleration vertically. The vertical motion can be described by the equation: \[ y = rac{1}{2} g t^2 \] where \(y\) is the vertical displacement (height below the bull's-eye), \(g\) is the acceleration due to gravity, and \(t\) is the time taken to reach the target.
02

Determine Time and Impact for Initial Shot

At the original distance from the target, the bullet falls a distance \( H_A \) in time \( t_1 \). Using horizontal motion, where the velocity is \( v_x \), the relation is: \[ ext{distance} = v_x t_1 \] For vertical motion: \[ H_A = \frac{1}{2} g t_1^2 \] This gives us the formula for initial fall due to gravity.
03

Extend Analysis to Double the Distance

When the rifle is fired again from twice the distance, the horizontal distance is \(2d = v_x t_2\), where \(t_2\) is the new time. Since the bullet travels twice the distance, \(t_2 = 2t_1\). Substituting into the vertical motion equation, the new height below the target is \[ H_B = \frac{1}{2} g (2t_1)^2 = 2^2 \times \frac{1}{2} g t_1^2 = 4 \times \frac{1}{2} g t_1^2 \] Which simplifies to \( H_B = 4H_A \).
04

Calculate the Ratio

The ratio of the distances \(H_B\) and \(H_A\) is:\[ \frac{H_B}{H_A} = \frac{4H_A}{H_A} = 4 \] Thus, the bullet falls four times the distance when the distance to the target is doubled.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravity Effects
In projectile motion, gravity plays a crucial role. It affects how quickly objects drop vertically when they are otherwise moving horizontally. For any object in freefall or moving along a curved trajectory, it experiences an acceleration due to gravity, represented as \( g \). This gravitational effect causes the bullet in the exercise to drop below the bull's-eye target, even though it is aimed directly at it from a horizontal perspective. Gravity accelerates all objects downwards at a constant rate, approximately \(9.81 \text{m/s}^2\) on Earth. Therefore, while the bullet travels horizontally towards the target, gravity pulls it downward incrementally. This is why in the equation \( y = \frac{1}{2} g t^2 \), \( y \) represents the vertical displacement, demonstrating how gravity increasingly influences the projectile over time. To predict or measure this vertical displacement, understanding and applying the concept of gravity's constant acceleration on an object is essential.
Horizontal and Vertical Motion
Projectile motion involves two distinct components: horizontal and vertical motion. Understanding how they work independently yet together helps solve problems like the bullet hitting below the bull's-eye.
  • Horizontal Motion: This is consistent and unaffected by gravity. If a bullet is fired from a rifle parallel to the ground, its horizontal velocity remains constant. This is because there is no horizontal force acting on it (assuming air resistance is negligible).
  • Vertical Motion: In contrast, vertical motion is greatly influenced by gravity. As the bullet travels forward, it also falls downward at an accelerating rate due to gravity.
The bullet's path is a combination of these two motions. While it moves steadily towards the target horizontally, it simultaneously drops towards the ground. This combined motion forms a curved path, which is a hallmark of projectile paths.
Time of Flight in Physics
Time of flight is the duration an object remains in motion from the moment it is projected until it hits the target. In the given exercise, understanding time of flight helps explain why doubling the distance results in the bullet falling four times as far. For horizontal motion, time of flight is crucial in determining how long the bullet travels towards the target at a given constant speed. As calculated in the original problem, if the target's distance is doubled, the time of horizontal flight doubles (\(t_2 = 2t_1\)), simply because the bullet covers more horizontal space.Vertically, the longer the time an object is in the air, the more it is affected by gravity. Therefore, if the time of flight increases, the gravitational pull causes a greater drop. This directly explains the increased vertical distance fallen (\( H_B = 4H_A \)) when the shooting distance doubles. Thus, understanding time of flight explains how variations in distance impact both the horizontal and vertical travel, giving a comprehensive picture of a projectile's motion.

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Most popular questions from this chapter

A small can is hanging from the ceiling. A rifle is aimed directly at the can, as the figure illustrates. At the instant the gun is fired, the can is released. Ignore air resistance and show that the bullet will always strike the can, regardless of the initial speed of the bullet. Assume that the bullet strikes the can before the can reaches the ground.

A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{~m} / \mathrm{s},\) due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{~m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{~m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

The captain of a plane wishes to proceed due west. The cruising speed of the plane is \(245 \mathrm{~m} / \mathrm{s}\) relative to the air. A weather report indicates that a \(38.0-\mathrm{m} / \mathrm{s}\) wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane relative to the air?

A volleyball is spiked so that it has an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) directed downward at an angle of \(55^{\circ}\) below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

As preparation for this problem, review Conceptual Example \(10 .\) The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of \(135 \mathrm{~m} / \mathrm{s}\), and each tank is at the Plane \(\mathrm{B}\) same height of \(2.00 \mathrm{~km}\) above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of \(15.0^{\circ}\) above the horizontal and the other is flying at an angle of \(15.0^{\circ}\) below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A and (b) plane B. In each part, give the directional angles with respect to the horizontal.

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