/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 A high-powered rifle fires a bul... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A high-powered rifle fires a bullet with a muzzle speed of \(1.00 \mathrm{km} / \mathrm{s} .\) The gun is pointed horizontally at a large bull's eye target-a set of concentric rings- \(200 \mathrm{m}\) away. (a) How far below the extended axis of the rifle barrel does a bullet hit the target? The rifle is equipped with a telescopic sight. It is "sighted in" by adjusting the axis of the telescope so that it points precisely at the location where the bullet hits the target at \(200 \mathrm{m}\). (b) Find the angle between the telescope axis and the rifle barrel axis. When shooting at a target at a distance other than \(200 \mathrm{m}\) the marksman uses the telescopic sight, placing its crosshairs to "aim high" or "aim low" to compensate for the different range. Should she aim high or low, and approximately how far from the bull's eye, when the target is at a distance of (c) \(50.0 \mathrm{m},\) (d) \(150 \mathrm{m},\) or (e) \(250 \mathrm{m} ?\) Note: The trajectory of the bullet is everywhere so nearly horizontal that it is a good approximation to model the bullet as fired horizontally in each case. What if the target is uphill or downhill? (f) Suppose the target is \(200 \mathrm{m}\) away, but the sight line to the target is above the horizontal by \(30^{\circ} .\) Should the marksman aim high, low, or right on? (g) Suppose the target is downhill by \(30^{\circ} .\) Should the marksman aim high, low, or right on? Explain your answers.

Short Answer

Expert verified
The bullet will hit 0.02m below the extended axis of the rifle barrel. The scope will form an angle of approximately \(1.146^{\circ}\) with the rifle barrel. When shooting at a target 50m away, aim high by approximately \(6 \times 10^{-4}m\), 150m away, aim high by approximately \(0.005m\), and at 250m, aim low by approximately \(0.015m\). If the target is uphill by \(30^{\circ}\), she should aim right on, and if it is downhill by \(30^{\circ}\), she should also aim right on.

Step by step solution

01

Calculate The Vertical Displacement

Using the standard equation for motion under gravity, the vertical displacement \(y\) covered by the bullet in time \(t\) is given by \(y = \frac{1}{2}gt^2\). The horizontal range \(x = vt\), thus the time \(t\) to target can be computed as \(t = \frac{x}{v}\), where \(v = 1km/s = 1000m/s\) and \(x = 200m\). Thus, substituting time into the equation for vertical displacement, we get \(y = \frac{1}{2}g(\frac{x}{v})^2\).
02

Calculate The Scope Angle

The angle \(\theta\) between the telescope axis and the rifle barrel axis can be computed using tan function of the ratio between vertical displacement \(y\) and horizontal distance \(x\), i.e. \(\theta = tan^{-1}(\frac{y}{x})\).
03

Aim Adjustment for Different Ranges

On analyzing the bullet’s trajectory, it is clear that for targets at less than 200m, the aim needs to be higher than the bull's eye and for targets more than 200m, the aim needs to be lowered. By using the displacement calculation as in Step 1, the deviation from the target can be computed to indicate how high or low to aim at 50m, 150m and 250m.
04

Uphill or Downhill Target

With reference to the tan function, if we add or remove angle component equivalent to uphill or downhill slope, we can accordingly decide whether to aim high or low. For uphill aim generally needs to be slightly higher according to the slope whereas for downhill aim has to be slightly lower.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Projectile Motion
Projectile motion describes the motion of an object that is thrown or projected into the air, subject to only the acceleration of gravity. The path that the object follows is known as its trajectory. In our scenario, a bullet fired from a high-powered rifle is considered a projectile. Since the rifle is aimed horizontally, the only force acting on the bullet once it leaves the barrel is gravity (ignoring air resistance).

When thinking about projectile motion, there are two key components to consider: horizontal and vertical motions. These two motions are independent of each other. The horizontal component is governed by the initial velocity of the bullet, which in the case of our exercise is a substantial 1 km/s. The bullet will maintain this horizontal speed throughout its flight because there are no horizontal forces acting on it after it's been fired. On the other hand, the vertical motion is governed by the gravity of Earth, pulling the projectile downward at an acceleration of approximately 9.81 m/s^2.
Motion Under Gravity
The term 'motion under gravity' refers to the movement of any object under the sole influence of gravity. This is a fundamental concept in physics, as it explains how objects behave when dropped, thrown, or launched horizontally, such as the bullet from our exercise. Under the influence of gravity, all objects will accelerate downward at the same rate, regardless of their mass.

This acceleration due to gravity, denoted by 'g', is constant near the Earth's surface and has a value of approximately 9.81 m/s^2. The downward path of the projectile is parabolic because the horizontal velocity is constant whereas the vertical velocity increases linearly with time due to the constant acceleration imparted by gravity.
Telescopic Sight Adjustment
In our exercise, the telescopic sight of a rifle is an essential tool for precision shooting. The primary function of the telescopic sight is to align the shooter's aim with the point of impact of the bullet. By 'sighting in' the scope, shooters can ensure their aim is accurate at a specific distance.

However, when the target distance changes, the shooter must adjust their aim because the bullet drops over longer distances due to gravity. This is where understanding the ballistic trajectory is crucial. At distances less than the sighted range, the bullet will impact above the line of sight, and so the shooter must aim lower ('aim high'). Conversely, at distances greater than the sighted range, the shooter must raise the aim point ('aim low') to compensate for the increased bullet drop.
Range of a Projectile
The range of a projectile is defined as the horizontal distance it travels before hitting the ground. For a bullet fired from a gun, the range is critical for marksmen to calculate where the bullet will fall in relation to the target. In a controlled environment, when firing a bullet horizontally at a known velocity, as in our exercise, the range can be predicted using the equations of projectile motion.

For a projectile fired at an angle, the range would depend on both the speed and angle of release. The range is longest when a projectile is fired at a 45-degree angle assuming a flat surface with no air resistance. However, in reality, air resistance, elevation changes, and other factors can significantly affect the range. Marksmen can use information about the range to make adjustments for targets at different distances by considering how much lower or higher they need to aim.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The water in a river flows uniformly at a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\) between parallel banks \(80.0 \mathrm{m}\) apart. You are to deliver a package directly across the river, but you can swim only at \(1.50 \mathrm{m} / \mathrm{s} .\) (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) What If? If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried?

A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of \(37.0^{\circ}\) below the horizontal. The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from rest at \(t=0,\) the car rolls down the incline with a constant acceleration of \(4.00 \mathrm{m} / \mathrm{s}^{2},\) traveling \(50.0 \mathrm{m}\) to the edge of a vertical cliff. The cliff is \(30.0 \mathrm{m}\) above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff and the time at which it arrives there, (b) the velocity of the car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the position of the car when it lands in the ocean, relative to the base of the cliff.

In the What If? section of Example 4.7 , it was claimed that the maximum range of a ski-jumper occurs for a launch angle \(\theta\) given by $$\theta=45^{\circ}-\frac{\phi}{2}$$ where \(\phi\) is the angle that the hill makes with the horizontal in Figure \(4.16 .\) Prove this claim by deriving the equation above.

A fisherman sets out upstream from Metaline Falls on the Pend Oreille River in northwestern Washington State. His small boat, powered by an outboard motor, travels at a constant speed \(v\) in still water. The water flows at a lower constant speed \(v_{w}\) He has traveled upstream for \(2.00 \mathrm{km}\) when his ice chest falls out of the boat. He notices that the chest is missing only after he has gone upstream for another 15.0 minutes. At that point he turns around and heads back downstream, all the time traveling at the same speed relative to the water. He catches up with the floating ice chest just as it is about to go over the falls at his starting point. How fast is the river flowing? Solve this problem in two ways. (a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels upstream at speed \(v-v_{w}\) and downstream at \(v+v_{w \cdot}\) (b) \(\mathrm{A}\) second much simpler and more elegant solution is obtained by using the water as the reference frame. This approach has important applications in many more complicated problems; examples are calculating the motion of rockets and satellites and analyzing the scattering of subatomic particles from massive targets. PICTURE CANT COPY

In a television picture tube (a cathode ray tube) electrons are emitted with velocity \(\mathbf{v}_{i}\) from a source at the origin of coordinates. The initial velocities of different electrons make different angles \(\theta\) with the \(x\) axis. As they move a distance \(D\) along the \(x\) axis, the electrons are acted on by a constant electric field, giving each a constant acceleration a in the \(x\) direction. At \(x=D\) the electrons pass through a circular aperture, oriented perpendicular to the \(x\) axis. At the aperture, the velocity imparted to the electrons by the electric field is much larger than \(\mathbf{v}_{i}\) in magnitude. Show that velocities of the electrons going through the aperture radiate from a certain point on the \(x\) axis, which is not the origin. Determine the location of this point. This point is called a virtual source, and it is important in determining where the electron beam hits the screen of the tube.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.