Q. 50 Projectile Motion The range R of... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 50 (page 501)

Projectile Motion The range R of a projectile propelled downward from the top of an inclined plane at an angle $胃$ to the inclined plane is given by
$R (胃) = \frac{2 v_{0}^{2} \sin 鈦� 胃 \cos 鈦� (胃鈭� 蠒)}{g \cos^{2} 鈦� 蠒}$
where $v_{o}$ is the initial velocity of the projectile, $鈭�$ is the angle the plane makes with respect to the horizontal, and g is the acceleration due to gravity.
(a) Show that for fixed $v_{o}$ and $鈭�$ the maximum range down the incline is given by
$R_{max} = \frac{v_{0}^{2}}{g (1 鈭� \sin 鈦� 蠒)}$
(b) Determine the maximum range if the projectile has an initial velocity of $50$ meters/second, the angle of the plane is $鈭� = 35^{o}$ , and $g = 9.8$ meters/second².

Short Answer

Expert verified

(a) $R_{m a x} = \frac{{v_{0}}^{2}}{g (1 鈭� \sin 鈦� 蠒)}$

(b) $598$ meters

Step by step solution

Part (a) Step 1. Maximum range

When $v_{0}$ is fixed along with $蠒$ , then the maximum range is given by,

$\begin{aligned} R (胃) = \frac{2 {v_{0}}^{2} \sin 鈦� 胃 \cos 鈦� (胃鈭� 蠒)}{g \cos^{2} 鈦� 蠒} \\ = \frac{2 {v_{0}}^{2} \frac{1}{2} [\sin 鈦� (胃 + 胃鈭� 蠒) + \sin 鈦� (胃鈭� (胃鈭� 蠒))]}{g \cos^{2} 鈦� 蠒} \\ = \frac{{v_{0}}^{2} [\sin 鈦� (2 胃鈭� 蠒) + \sin 鈦� 蠒]}{g \cos^{2} 鈦� 蠒} \end{aligned}$

and,

role="math" localid="1646563360195" $\begin{aligned} \sin 鈦� (2 胃鈭� 蠒) = 1 \\ \frac{v_{0}^{2} (1 + \sin 鈦� 蠒)}{g \cos^{2} 鈦� 蠒} = \frac{{v_{0}}^{2} [1 + \sin 鈦� 蠒]}{g (1 鈭� \sin^{2} 鈦� 蠒)} \\ = \frac{{v_{0}}^{2} (1 + \sin 鈦� 蠒)}{g (1 + \sin 鈦� 蠒) (1 鈭� \sin 鈦� 蠒)} \\ = \frac{{v_{0}}^{2}}{g (1 鈭� \sin 鈦� 蠒)} \end{aligned}$

Part (b) Step 1. Maximum range

Let us consider that,

$v_{0} = 50 蠒 = 35^{掳} g = 9.8$

So the maximum range is given by,

$\begin{aligned} R_{max} = \frac{{v_{0}}^{2}}{g (1 鈭� \sin 鈦� 蠒)} \\ = \frac{50^{2}}{9.8 (1 鈭� \sin 鈦� 35^{鈭�})} \\ = 598 meters \end{aligned}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Part (a) Step 1. Maximum range

Part (b) Step 1. Maximum range

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Theoretical and Mathematical Physics

Decision Maths

Applied Mathematics

Geometry

Mechanics Maths

Study anywhere. Anytime. Across all devices.