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,

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

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91影视

Short Answer

Step by step solution

Part (a) Step 1. Maximum range

Part (b) Step 1. Maximum range

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