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Chapter 13: Problem 29

Equal-range firing angles enable a projectile to reach a target 16 \(\mathrm{km}\) downrange on the same level as the gun if the projectile's initial speed is 400 \(\mathrm{m} / \mathrm{sec}\) ?

Short Answer

Expert verified
The projectile can be launched at angles approximately \( 39.39^\circ \) and \( 50.61^\circ \).

Step by step solution

01

Define the Range Formula for Projectile Motion

For projectile motion, the range \( R \) is given by the formula \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( v_0 \) is the initial speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \).
02

Substitute Known Values into the Formula

Substitute the given values into the equation: \( v_0 = 400 \, \text{m/s} \) and \( R = 16000 \, \text{m} \). The equation becomes \( 16000 = \frac{400^2 \sin(2\theta)}{9.81} \).
03

Solve for \( \sin(2\theta) \)

Rearrange the equation to solve for \( \sin(2\theta) \): \[ \sin(2\theta) = \frac{16000 \times 9.81}{400^2} \]. Calculate this value to find \( \sin(2\theta) \approx 0.981 \).
04

Find Possible Angles for \( \theta \)

Using the value \( \sin(2\theta) \approx 0.981 \), find the possible angles for \( 2\theta \). Thus, \( 2\theta = \sin^{-1}(0.981) \), which gives \( 2\theta \approx 78.79^\circ \) and \( 2\theta \approx 101.21^\circ \).
05

Determine Equal-Range Angles

Since \( 2\theta_1 = 78.79^\circ \), \( \theta_1 \approx 39.39^\circ \). For \( 2\theta_2 = 101.21^\circ \), subtract from \( 180^\circ \): \( \theta_2 = 90^\circ - \theta_1 \approx 50.61^\circ \). Therefore, the launch angles are approximately \( 39.39^\circ \) and \( 50.61^\circ \).

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Key Concepts

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

Range Formula
In projectile motion, when we talk about how far a projectile will travel horizontally before hitting the ground, we're referring to the *range*. This is given by the range formula: \( R = \frac{v_0^2 \sin(2\theta)}{g} \) Here:
  • \( R \) represents the range or horizontal distance.
  • \( v_0 \) is the initial speed of the projectile.
  • \( \theta \) stands for the launch angle, the angle at which the projectile is released.
  • \( g \) is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \) on Earth.
This formula takes into account all these factors to predict how far an object will travel based solely on its initial kick-off speed and angle. It's crucial in physics because it gives a comprehensive view tied to practical scenarios, like launching a projectile over a certain distance.
Launch Angle
The launch angle, denoted as \( \theta \), is the angle at which a projectile is fired or launched with respect to the horizontal axis. It's an essential element in determining how far and at what trajectory a projectile will travel. Different launch angles will affect the flight path and the range of the projectile.
  • A launch angle of \( 45^\circ \) is often considered ideal for maximizing range when air resistance is negligible.
  • Launch angles that are complementary (e.g., \( \theta \) and \( 90^\circ - \theta \) ) result in the same range.
In the problem discussed, two different launch angles, approximately \( 39.39^\circ \) and \( 50.61^\circ \), were calculated, emphasizing how different angles can achieve the same downrange target.
Initial Speed
Initial speed, represented as \( v_0 \), is the velocity at which a projectile is launched. In the context of projectile motion, this speed significantly influences how far and fast the projectile travels. The initial speed acts as an input to the range formula and has a substantial effect on the range when combined with the launch angle.
  • In this specific problem, the initial speed was given as \( 400 \, \text{m/s} \).
  • The speed was critical in reaching the target that was \( 16 \, \text{km} \) away.
  • Higher initial speeds generally result in longer travel distances, assuming the other factors remain constant.
Understanding initial speed is key to solving projectile motion problems correctly, as it is directly tied into the equations that dictate how projectiles move through space.
Acceleration Due to Gravity
The acceleration due to gravity, commonly represented as \( g \), plays a pivotal role in projectile motion. On Earth, this acceleration is about \( 9.81 \, \text{m/s}^2 \), pulling objects toward the surface. This force affects all aspects of a projectile's motion, such as its trajectory and the time it remains airborne.
  • Gravity acts downwards, consistently modifying the path of any projectile.
  • In determining range, \( g \) acts as a divisor in the range formula \( R = \frac{v_0^2 \sin(2\theta)}{g} \). The smaller \( g \), the greater the range for a given initial speed and launch angle.
Considering gravity is crucial, as it helps in predicting the motion of a projectile accurately by accounting for the downward pull that impacts its flight.

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Most popular questions from this chapter

Flight time and height A projectile is fired with an initial speed of 500 \(\mathrm{m} / \mathrm{sec}\) at an angle of elevation of \(45^{\circ} .\) a. When and how far away will the projectile strike? b. How high overhead will the projectile be when it is 5 \(\mathrm{km}\) downrange? c. What is the greatest height reached by the projectile?

A formula for the curvature of a parametrized plane curve $$ \begin{array}{c}{\text { a. Show that the curvature of a smooth curve } \mathbf{r}(t)=f(t) \mathbf{i}+} \\ {g(t) \mathbf{j} \text { defined by twice- differentiable functions } x=f(t) \text { and }} \\ {y=g(t) \text { is given by the formula }} \\ {\kappa=\frac{|\ddot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}}}\end{array} $$The dots in the formula denote differentiation with respect to \(t,\) one derivative for each dot. Apply the formula to find the curvatures of the following curves. $$ \begin{array}{l}{\text { b. } \mathbf{r}(t)=t \mathbf{i}+(\ln \sin t) \mathbf{j}, \quad 0 < t<\pi} \\ {\text { c. } \mathbf{r}(t)=\left[\tan ^{-1}(\sinh t)\right] \mathbf{i}+(\ln \cosh t) \mathbf{j}}\end{array} $$

Write a in the form \(\mathbf{a}=a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) at the given value of \(t\) without finding \(\mathbf{T}\) and \(\mathbf{N} .\) \(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t+(1 / 3) t^{3}\right) \mathbf{j}+\left(t-(1 / 3) t^{3}\right) \mathbf{k}, \quad t=0\)

Linear drag Derive the equations $$\begin{aligned} x &=\frac{v_{0}}{k}\left(1-e^{-k t}\right) \cos \alpha \\ y &=\frac{v_{0}}{k}\left(1-e^{-k t}\right)(\sin \alpha)+\frac{g}{k^{2}}\left(1-k t-e^{-k t}\right) \end{aligned}$$ by solving the following initial value problem for a vector \(r\) in the plane. Differential equation: $$\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}$$ Initial conditions:$$\mathbf{r}(0)=\mathbf{0}\( \)\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=\mathbf{v}_{0}=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}$$ The drag coefficient \(k\) is a positive constant representing resistance due to air density, \(v_{0}\) and \(\alpha\) are the projectile's initial speed and launch angle, and \(g\) is the acceleration of gravity.

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{c}{\text { Differential equation: }} \\ {\frac{d \mathbf{r}}{d t}=\left(\frac{t}{t^{2}+2}\right) \mathbf{i}-\left(\frac{t^{2}+1}{t-2}\right) \mathbf{j}+\left(\frac{t^{2}+4}{t^{2}+3}\right) \mathbf{k}} \\ {\text { Initial condition: } \mathbf{r}(0)=\mathbf{i}-\mathbf{j}+\mathbf{k}}\end{array} $$
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