A projectile fired from the ground follows the trajectory given by the following equation, where is the initial speed, is the angle of projection, g is the acceleration due to gravity, and k is the drag factor caused by air resistance. y=tan(0) + y = (tan(0))x - y =tan(0) + x² Using the power series representation In(1 + x) = x- 2 y =tan(0)x - g kvo cos(8) 05 (0) ) x + 2/2 In (1- k² 9 kvo cos(8) 2 Σ n = 2 gx² ² cos²(0) 2V0 √5 ) x + ²/2² ( 2₁ kx Vo cos(8) 3vo + kgx³ ³ cos³(0) x3 x4 3 4 ..., -1 < x < 1, verify that the trajectory can be rewritten as the following. k²gx4 4 cos (0) 4v0
A projectile fired from the ground follows the trajectory given by the following equation, where is the initial speed, is the angle of projection, g is the acceleration due to gravity, and k is the drag factor caused by air resistance. y=tan(0) + y = (tan(0))x - y =tan(0) + x² Using the power series representation In(1 + x) = x- 2 y =tan(0)x - g kvo cos(8) 05 (0) ) x + 2/2 In (1- k² 9 kvo cos(8) 2 Σ n = 2 gx² ² cos²(0) 2V0 √5 ) x + ²/2² ( 2₁ kx Vo cos(8) 3vo + kgx³ ³ cos³(0) x3 x4 3 4 ..., -1 < x < 1, verify that the trajectory can be rewritten as the following. k²gx4 4 cos (0) 4v0
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![A projectile fired from the ground follows the trajectory given by the following equation, where \( v_0 \) is the initial speed, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, and \( k \) is the drag factor caused by air resistance.
\[
y = \left( \tan(\theta) + \frac{g}{kv_0 \cos(\theta)} \right)x + \frac{g}{k^2} \ln \left( 1 - \frac{kx}{v_0 \cos(\theta)} \right)
\]
Using the power series representation \(\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots\), \(-1 < x < 1\), verify that the trajectory can be rewritten as the following.
\[
y = (\tan(\theta))x - \frac{gx^2}{2v_0^2 \cos^2(\theta)} - \frac{kgx^3}{3v_0^3 \cos^3(\theta)} - \frac{k^2 gx^4}{4v_0^4 \cos^4(\theta)} - \cdots
\]
\[
y = \left( \tan(\theta) + \frac{g}{kv_0 \cos(\theta)} \right)x + \frac{g}{k^2} \left( \sum_{n = 1}^{\infty} \_\_\_\_\_ \right)
\]
\[
y = \tan(\theta)x - \sum_{n = 2}^{\infty} \_\_\_\_\_
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8530ca5f-6e8e-4cce-aaa4-95b39b1a4e45%2Fe4c126d6-1c03-4c04-9c34-44feed15cae7%2Frot827u_processed.png&w=3840&q=75)
Transcribed Image Text:A projectile fired from the ground follows the trajectory given by the following equation, where \( v_0 \) is the initial speed, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, and \( k \) is the drag factor caused by air resistance.
\[
y = \left( \tan(\theta) + \frac{g}{kv_0 \cos(\theta)} \right)x + \frac{g}{k^2} \ln \left( 1 - \frac{kx}{v_0 \cos(\theta)} \right)
\]
Using the power series representation \(\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots\), \(-1 < x < 1\), verify that the trajectory can be rewritten as the following.
\[
y = (\tan(\theta))x - \frac{gx^2}{2v_0^2 \cos^2(\theta)} - \frac{kgx^3}{3v_0^3 \cos^3(\theta)} - \frac{k^2 gx^4}{4v_0^4 \cos^4(\theta)} - \cdots
\]
\[
y = \left( \tan(\theta) + \frac{g}{kv_0 \cos(\theta)} \right)x + \frac{g}{k^2} \left( \sum_{n = 1}^{\infty} \_\_\_\_\_ \right)
\]
\[
y = \tan(\theta)x - \sum_{n = 2}^{\infty} \_\_\_\_\_
\]
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