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According to Google, the average initial velocity of a typical cannonball is 820 miles per hour. Ballistics theory indicates that the optimal angle to fire a projectile to maximize distance is 45°. Under these conditions, how far would a cannonball travel and how long would it take to hit the ground?

According to Google, the average initial velocity of a typical cannonball is 820 miles per hour. Ballistics theory indicates that the optimal angle to fire a projectile to maximize distance is 45°. Under these conditions, how far would a cannonball travel and how long would it take to hit the ground?

Calculus: Early Transcendentals
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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 Use the technique described in example 5 to solve the following problem.

- Example 5 is attached 

According to Google, the average initial velocity of a typical cannonball is 820 miles per hour. Ballistics theory indicates that the optimal angle to fire a projectile to maximize distance is 45°. Under these conditions, how far would a cannonball travel and how long would it take to hit the ground?

is the example. Applications EXAMPLE 5 Suppose a projectile is fired at an angle of elevation 0 with an initial velocity of vo ft/sec, as illustrated in Figure 18. In calculus and physics courses you will learn that points (x(t), y(t)) on the path of the projectile at time t are given by x(t) = (vo cos 0)t and y(t) = (vo sin @)t - zg, 400 ft/sec 30 FIGURE 18 where g 32ft/sec? is the acceleration due to the earth's gravity. Where will a pro- jectile land if it is fired at an angle of 30° and an initial velocity of 400 ft/sec? Solution For the given values of 0 and vo we have V3 x(t) = (400 cos 30°)t = 400– 200/3t, t = and y(t) = (400 sin 30°)t = 400 G)t- 167? = 200t – 167. %3D The projectile hits the ground at the value of t > 0 for which y(t) = 0. This occurs when 0 = 200t – 167 = 8t(25 – 2t), t = 12.5 seconds. SO Copyright 2010 Cangga Learning All Rights Ravaned. May not be copied scanned, or duplicated in whole or in part. Due to electronic rights, sce fhird party contant may be suppressed from the aBook and'or eChaptar(s). Editorial reniow has doamed that aury supprezed contaat does not materially affect the overall learning experienca. Cangage Learning reverves the right to remore additional contert at zuy time if subvequant rights restrictions requira it. 4.4 The Sine and Cosine Functions 219 Hence the projectile hits the ground x(12.5) = 200/3(12.5) z 200(1.732)(12.5) × 1169 feet from where it was fired.
Transcribed Image Text:is the example. Applications EXAMPLE 5 Suppose a projectile is fired at an angle of elevation 0 with an initial velocity of vo ft/sec, as illustrated in Figure 18. In calculus and physics courses you will learn that points (x(t), y(t)) on the path of the projectile at time t are given by x(t) = (vo cos 0)t and y(t) = (vo sin @)t - zg, 400 ft/sec 30 FIGURE 18 where g 32ft/sec? is the acceleration due to the earth's gravity. Where will a pro- jectile land if it is fired at an angle of 30° and an initial velocity of 400 ft/sec? Solution For the given values of 0 and vo we have V3 x(t) = (400 cos 30°)t = 400– 200/3t, t = and y(t) = (400 sin 30°)t = 400 G)t- 167? = 200t – 167. %3D The projectile hits the ground at the value of t > 0 for which y(t) = 0. This occurs when 0 = 200t – 167 = 8t(25 – 2t), t = 12.5 seconds. SO Copyright 2010 Cangga Learning All Rights Ravaned. May not be copied scanned, or duplicated in whole or in part. Due to electronic rights, sce fhird party contant may be suppressed from the aBook and'or eChaptar(s). Editorial reniow has doamed that aury supprezed contaat does not materially affect the overall learning experienca. Cangage Learning reverves the right to remore additional contert at zuy time if subvequant rights restrictions requira it. 4.4 The Sine and Cosine Functions 219 Hence the projectile hits the ground x(12.5) = 200/3(12.5) z 200(1.732)(12.5) × 1169 feet from where it was fired.
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