/*! 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 35 A rifle that shoots bullets at \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Short Answer

Expert verified
Aim the rifle 4.8 cm above the target to hit dead center.

Step by step solution

01

Identify the Known Variables

We know the initial velocity \( v_0 = 460 \mathrm{~m/s} \) and the horizontal distance to the target \( d = 45.7 \mathrm{~m} \). We'll also use acceleration due to gravity \( g = 9.81 \mathrm{~m/s}^2 \).
02

Calculate Time of Flight

The time \( t \) the bullet takes to reach the target can be calculated using the formula for horizontal motion: \( d = v_0 \cos(\theta) \cdot t \). Since the target is level with the rifle, \( \theta = 0 \) implies \( v_0 \cos(\theta) = v_0 \). Thus, \( t = \frac{d}{v_0} = \frac{45.7}{460} \approx 0.0993 \mathrm{~s} \).
03

Calculate the Vertical Drop

During the time of flight, the bullet will drop due to gravity. The vertical drop \( h \) can be calculated with \( h = \frac{1}{2}gt^2 \). Plugging the values gives \( h = \frac{1}{2} \times 9.81 \times (0.0993)^2 \approx 0.048 \mathrm{~m} \).
04

Determine the Required Height Above Target

For the bullet to hit the dead center, the rifle must be aimed above the target to compensate for the vertical drop. Therefore, the rifle must be pointed \( 0.048 \mathrm{~m} \) or \( 4.8 \mathrm{~cm} \) above the target center.

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.

Horizontal Motion
Horizontal motion in projectile motion refers to the motion of an object in a straight path parallel to the ground. In the given exercise, the bullet fired from the rifle travels horizontally towards the target. This motion is characterized by a constant horizontal velocity since there are no horizontal forces acting on the bullet (assuming air resistance is negligible).

To determine how far the bullet travels horizontally, we use the simple distance formula:
  • Distance ( \(d\)) = Horizontal Velocity ( \(v_{0x}\)) \(\times\) Time ( \(t\))
The initial horizontal velocity \( v_{0x}\) is given as \(460\, \mathrm{m/s}\). Since the horizontal velocity remains constant, the time taken to reach the target can be calculated by dividing the horizontal distance \( \mathrm{d}\) by \(v_{0x}\).

For this exercise, the horizontal distance to the target was \(45.7\, \mathrm{m}\), and the time taken was calculated to be approximately \(0.0993\, \mathrm{s}\). The constancy of horizontal motion helps predict where the projectile will land along the horizontal axis.
Vertical Motion
Unlike horizontal motion, vertical motion in a projectile follows a parabolic path and is affected by gravity. This component of motion is crucial to calculating how much a projectile drops while moving horizontally. In our scenario with the rifle, the vertical motion determines how far the bullet falls before it hits the target.

The vertical displacement or drop \( h \) of the bullet is due to the acceleration from gravity. It can be found using the formula:
  • \[ h = \frac{1}{2}gt^2 \]
Where:
  • \( g \) is the acceleration due to gravity \(\approx 9.81 \, \mathrm{m/s^2}\).
  • \( t \) is the time of flight calculated from horizontal motion.
For this exercise, the bullet's vertical drop was found to be \(0.048\, \mathrm{m}\) or \(4.8\, \mathrm{cm}\). The vertical motion needs to be considered because this drop informs how much higher the shooter must aim above the target to ensure the projectile reaches the center once gravity acts on it.
Acceleration due to Gravity
Acceleration due to gravity is a constant force that acts on objects in free fall near the Earth's surface. Its value is approximately \(9.81\, \mathrm{m/s^2}\). This universal force affects any projectile's vertical motion, pulling it downward as it travels horizontally.

In the context of projectile motion, understanding gravity's role is crucial for calculating how high above a target a projectile, like a bullet, will drop during its flight. This includes:
  • Determining the time it takes for the bullet to reach the target based on horizontal distance.
  • Calculating the vertical drop that occurs within that time using the formula:
\[ h = \frac{1}{2}gt^2 \]

Gravity's influence means that even with a powerful velocity of \(460\, \mathrm{m/s}\), a bullet will still drop over the short span it travels to a target \(45.7\, \mathrm{m}\) away. This drop necessitates aiming higher to counteract gravity's effect, ensuring accuracy in hitting the intended target point.

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

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(4.0 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0 \mathrm{~m} / \mathrm{s}\), what is the smallest angle it can have for the pass to be legal?

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

A proton initially has \(\vec{v}=4.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and then \(4.0 \mathrm{~s}\) later has \(\vec{v}=-2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+5.0 \hat{\mathrm{k}}\) (in meters per second). For that \(4.0 \mathrm{~s}\), what are (a) the proton's average acceleration \(\vec{a}_{\text {avg }}\) in unitvector notation, (b) the magnitude of \(\vec{a}_{\text {avg }}\), and \((\) c) the angle between \(\vec{a}_{\text {avg }}\) and the positive direction of the \(x\) axis?

Suppose that a space probe can withstand the stresses of a \(20 g\) acceleration. (a) What is the minimum turning radius of such a craft moving at a speed of one-tenth the speed of light? (b) How long would it take to complete a \(90^{\circ}\) turn at this speed?

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=40.0^{\circ}\) above the horizontal (Fig. 4-35). The wall is distance \(d=\) \(22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

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.