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A rifle is aimed horizontally at a target \(30 \mathrm{~m}\) away. 'The bullet hits the target \(1.9 \mathrm{~cm}\) below the aiming point. What are (a) the bullet's time of flight and (b) its speed as it emerges from the rifle?

Short Answer

Expert verified
The bullet's time of flight is approximately 0.062 s, and its speed is about 484 m/s.

Step by step solution

01

Identify the known variables

We know the horizontal distance to the target, which is \(d = 30 \text{ m}\). The vertical distance the bullet falls is \(h = 1.9 \text{ cm} = 0.019 \text{ m}\). The gravitational acceleration is \(g = 9.81 \text{ m/s}^2\).
02

Calculate time of flight using vertical motion

The vertical motion of the bullet can be described by the equation \(h = \frac{1}{2} g t^2\). Solving for \(t\), we have: \[ t = \sqrt{\frac{2h}{g}} \]Substitute the known values:\[ t = \sqrt{\frac{2 \times 0.019}{9.81}} \approx 0.062 \text{ s} \]
03

Calculate bullet speed from horizontal motion

The horizontal motion is constant velocity motion, where the speed \(v\) is given by the equation \(v = \frac{d}{t}\). Substitute the known values:\[ v = \frac{30}{0.062} \approx 484 \text{ m/s} \]
04

Verify units and check calculations

Ensure that time is in seconds, distance in meters, and velocity in meters per second throughout the calculations. Re-calculate if any values seem incorrect.

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Key Concepts

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

Horizontal Motion
Horizontal motion describes how an object moves along the horizontal axis, parallel to the ground. For projectile motion, horizontal motion is often straightforward because air resistance is typically negligible in simple physics problems. This means the velocity stays constant from the moment it leaves the projectile (in this case, the rifle) until it reaches its target.

Understanding the horizontal motion is crucial because it allows us to calculate important characteristics of projectile motion, such as time of flight and horizontal distance.
  • In our problem, the bullet travels a horizontal distance of 30 meters.
  • Since we know the time it takes, we can find the speed simply by dividing distance by time.
Here, the horizontal motion is independent of vertical motion. They do not affect each other, which allows us to analyze them separately.
Vertical Motion
Vertical motion in projectile problems is influenced heavily by gravity. Unlike horizontal motion, vertical motion changes over time due to gravitational acceleration. In our example, the bullet drops 1.9 cm below the aiming point because of gravity.

Key aspects of vertical motion involve the distance fallen and the time taken to fall. You can calculate how far an object will fall using the formula:
  • Height fallen, \( h = \frac{1}{2} g t^2 \)
This equation shows that the falling distance is proportional to the square of the time elapsed. It highlights how gravity causes the bullet to drop more quickly as time increases. Calculating vertical motion requires understanding how gravity operates on an object.
Equations of Motion
Equations of motion form the foundation for analyzing projectile motion problems. These equations help us calculate various parameters such as displacement, velocity, and time. There are three main kinematic equations, but in projectile problems, two are most prominent:

1. The vertical motion equation:
  • \( h = \frac{1}{2} g t^2 \)
This equation is used to calculate how far something drops under gravity over time.

2. The horizontal motion equation:
  • \( v = \frac{d}{t} \)
This one is used to determine the constant speed required to travel a certain distance over a given period. In both equations, understanding how variables like time \( t \) and acceleration \( g \) interplay is essential. After determining time using vertical motion, you can figure out horizontal velocity using these equations.
Gravitational Acceleration
Gravitational acceleration is a key component in projectile motion, particularly in vertical motion. It represents how quickly an object accelerates towards the Earth due to gravity alone. This value is standardized at approximately \( 9.81 \text{ m/s}^2 \), influencing every object that moves vertically.

Gravitational acceleration causes the bullet in our problem to fall from its initial trajectory. It's crucial to understand that this acceleration is constant and acts on the object throughout its motion.

This consistent force allows us to predict the vertical motion accurately using the kinematic equations. Without gravitational acceleration, vertical displacement wouldn't occur the same way, demonstrating its importance in calculations involving vertical motion.

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Most popular questions from this chapter

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} \cdot\) At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis? 1

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Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(53 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

The position vector for an clectron is \(\vec{r}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((3.0 \mathrm{~m}) \hat{\mathrm{j}}+(2.0 \mathrm{~m}) \hat{\mathrm{k}},(\mathrm{a})\) Find the magnitude of \(\vec{r} .\) (b) Sketch the vector on a right-handed coordinate system.

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