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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 barrel 0.048 m above the target center.

Step by step solution

01

Understand the Problem

We need to determine how high above the center of the target the rifle barrel must be aimed so that the bullet hits the target exactly at the center. Given data includes the horizontal speed of the bullet, the distance to the target, and the fact that the target is level with the rifle barrel.
02

Determine Time of Flight

The bullet is traveling horizontally at a velocity of \(460 \, \text{m/s}\) and needs to cover a horizontal distance of \(45.7 \, \text{m}\). We calculate the time \( t \) it takes for the bullet to reach the target using the formula for time: \[ t = \frac{d}{v} = \frac{45.7}{460} \approx 0.0993 \text{ s} \]
03

Use the Vertical Motion Equation

Since the rifle is level with the target, the bullet will experience free fall, moving vertically downward due to gravity while it travels horizontally. Using the vertical motion equation \( y = \frac{1}{2}gt^2\), where \( g = 9.8 \text{ m/s}^2\), we can compute the vertical distance \( y \) the bullet drops during the flight time: \[ y = \frac{1}{2} \times 9.8 \times (0.0993)^2 \approx 0.048 \text{ m} \]
04

Calculate the Required Barrel Elevation

Since the bullet drops \(0.048 \text{ m}\) (or \(4.8 \text{ cm}\)) below the center due to gravity, the rifle barrel must be pointed \(0.048 \text{ m}\) higher than the target center to compensate for the drop, ensuring the bullet hits dead center.

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Key Concepts

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

Rifle Aiming
When aiming a rifle, particularly in physics problems, it is essential to consider both the horizontal and vertical components of projectile motion.
A rifle barrel must be aimed with precision to ensure that the bullet reaches the desired point on the target.
Understanding the mechanics involved here helps us appreciate why slight adjustments are necessary:
  • Horizontal aiming usually involves aligning the barrel with the target based on line of sight.
  • Vertical aiming must account for the effects of gravity on the bullet's trajectory.
In the case of a level target, the barrel must be elevated to balance the gravitational pull that causes a bullet to drop as it travels.
Proper aiming adjustments ensure that the bullet's path compensates for the vertical descent during its flight.
This is why we calculate how high above the target's center the rifle must be pointed.
Understanding this principle allows a precise hit at the intended mark.
Bullet Trajectory
Bullet trajectory refers to the path a bullet takes once it has been fired from a rifle.
It's a complex motion that encompasses both horizontal motion and a vertical descent due to gravity.
Gravity, a constant force pulling downward, influences the trajectory as follows:
  • The bullet travels horizontally to cover the distance to the target.
  • Simultaneously, it falls vertically, affected by gravity, meaning it drops from the line of sight.
The horizontal velocity remains constant in this scenario because there is no horizontal acceleration.
Meanwhile, the vertical motion can be described by the equation:
\( y = \frac{1}{2}gt^2 \)
where \( g \) represents gravitational acceleration, approximately \( 9.8 \, \text{m/s}^2 \).
The time \( t \) is derived from the horizontal motion formula, given a consistent horizontal speed.
By calculating both components, you can accurately predict where the bullet will land, allowing for precise targeting.
Physics Problem Solving
Solving physics problems, like determining the bullet trajectory of a rifle, involves a methodical approach blending physics principles and mathematical calculation.
Breaking down problems into manageable steps is crucial:
  • First, understand the scenario and identify what needs calculating. Here, it's the height difference for the rifle barrel aiming.
  • Next, determine what you know: the bullet's speed, the distance, and the absence of elevation between muzzle and target.
  • Calculate the time for the bullet to travel the distance using \( t = \frac{d}{v} \).
  • Apply vertical motion equations, recognizing that gravity affects the trajectory.
By integrating these steps effectively, you can comprehend multiple forces at play and how they influence the outcome.
This systematic approach not only helps in textbook problems but also enhances problem-solving skills in real-world scenarios.
Remember, mastering physics is about practice and understanding the core principles integrated into solving each unique problem.

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Most popular questions from this chapter

A 200 -m-wide river has a uniform flow speed of \(1.1 \mathrm{~m} / \mathrm{s}\) through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of \(4.0 \mathrm{~m} / \mathrm{s}\) with respect to the water. There is a clearing on the north bank \(82 \mathrm{~m}\) upstream from a point directly opposite the clearing on the south bank. (a) In what direction must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

In \(3.50 \mathrm{~h}\), a balloon drifts \(21.5 \mathrm{~km}\) north, \(9.70 \mathrm{~km}\) east, and \(2.88 \mathrm{~km}\) upward from its release point on the ground. Find (a) the magnitude of its average velocity and (b) the angle its average velocity makes with the horizontal.

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flagpole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0\) s later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0\) s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (c) magnitude and (f) direction of the average acceleration?

An elevator without a ceiling is ascending with a constant speed of \(10 \mathrm{~m} / \mathrm{s}\). A boy on the elevator shoots a ball directly upward, from a height of \(2.0 \mathrm{~m}\) above the elevator floor, just as the elevator floor is \(28 \mathrm{~m}\) above the ground. The initial speed of the ball with respect to the elevator is \(20 \mathrm{~m} / \mathrm{s}\). (a) What maximum height above the ground does the ball reach? (b) How long does the ball take to return to the elevator floor?

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

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