Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 214: 59

Answer

a) $\theta_{max}=\frac{1}{2}\tan ^{-1}\left(\dfrac{ mg }{ F_{wind} }\right)$ b) $11.5\%$

Work Step by Step

First, we need to draw the force diagram exerted on the projectile while there is an air resistance force exerted on it horizontally, as we see in the figure below. Now we need to apply Newton's second law on the projectile; $$\sum F_x=-F_{wind}=ma_x$$ Thus, $$a_x=\dfrac{-F_{wind}}{m}\tag 1$$ Now we need to find the time of the trip; $$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$ where $y=y_0=0$ since it starts from the ground and ends on the ground as well. Noting that $a_y=-g$ and $v_{0y}=v_0\sin\theta$. Thus, $$0=0+v_0\sin\theta t-\frac{1}{2}gt^2$$ $$0=t\left( v_0\sin\theta -\frac{1}{2}gt\right)$$ whether $t=0$ which is when the projectile fired, or $$t=\dfrac{2 v_0\sin\theta }{g}\tag 2$$ Now we need to find the range; $$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$$ where $v_{0x}=v_0\cos\theta$, $x-x_0=R$, and $a_x$ from (1); $$R= v_0\cos\theta t+\dfrac{-F_{wind}}{2m}t^2$$ Plugging $t$ from (2); $$R= v_0\cos\theta \left[\dfrac{2 v_0\sin\theta }{g}\right]-\dfrac{ F_{wind}}{2m} \left[\dfrac{2 v_0\sin\theta }{g}\right]^2$$ $$R= \left[\dfrac{2 v_0^2\sin\theta\cos\theta }{g}\right]-\dfrac{ F_{wind}}{2m} \left[\dfrac{4 v_0^2\sin^2\theta }{g^2}\right] $$ Noting that $\sin2\theta=2\sin\theta\cos\theta$ $$R= \dfrac{ v_0^2\sin2\theta }{g} - \dfrac{2 F_{wind} v_0^2\sin^2\theta }{ mg^2} $$ $$R= \left( \dfrac{ v_0^2 }{g} \right)\sin2\theta- \left(\dfrac{2 F_{wind} v_0^2 }{ mg^2} \right) \sin^2\theta\tag 3$$ To maximize the range, we need to take the first derivative relative to $\theta$ equal to zero. $$\dfrac{dR}{d\theta}=0= \dfrac{d }{d\theta}\left( \dfrac{ v_0^2 }{g} \right)\sin2\theta- \dfrac{d }{d\theta} \left(\dfrac{2 F_{wind} v_0^2 }{ mg^2} \right) \sin^2\theta$$ $$0= \left( \dfrac{ v_0^2 }{g} \right)2\cos2\theta- \left(\dfrac{2 F_{wind} v_0^2 }{ mg^2} \right) 2\sin \theta\cos\theta$$ $$0= \left( \dfrac{ v_0^2 }{g} \right)2\cos2\theta- \left(\dfrac{2 F_{wind} v_0^2 }{ mg^2} \right) \sin2 \theta $$ $$ \left( \dfrac{ \color{red}{\bf\not} v_0^2 }{\color{red}{\bf\not} g} \right)\color{red}{\bf\not} 2\cos2\theta= \left(\dfrac{\color{red}{\bf\not} 2 F_{wind} \color{red}{\bf\not} v_0^2 }{ mg^{\color{red}{\bf\not} 2}} \right) \sin2 \theta $$ $$ \cos2\theta= \left(\dfrac{ F_{wind} }{ mg } \right) \sin2 \theta $$ Hence, $$\tan 2\theta=\dfrac{ mg }{ F_{wind} }$$ $$2\theta=\tan ^{-1}\left(\dfrac{ mg }{ F_{wind} }\right)$$ Therefore, $$ \boxed{\theta_{max}=\frac{1}{2}\tan ^{-1}\left(\dfrac{ mg }{ F_{wind} }\right)}$$ --- b) First we need to find the maximum range when air resistance is negligible. $$R_1=\dfrac{v_0^2\sin 2\theta}{g}$$ where $\theta=45^\circ$ when there is no air friction. $$R_1=\dfrac{v_0^2\sin (2\times45^\circ)}{g}$$ $$R_1=\dfrac{v_0^2 }{g}\tag 4$$ Now we need to find the maximum range when we have air resistance where the angle is given by $$ \theta' =\frac{1}{2}\tan ^{-1}\left(\dfrac{ mg }{ F_{wind} }\right)= \frac{1}{2}\tan ^{-1}\left(\dfrac{ 0.5\times 9.8}{ 0.6 }\right) $$ $$\theta'=\bf 41.51^\circ $$ Now we need to find $R_2$ which is given by (3); $$R_2= \left( \dfrac{ v_0^2 }{g} \right)\sin2\theta'- \left(\dfrac{2 F_{wind} v_0^2 }{ mg^2} \right) \sin^2\theta'$$ Plugging the known; $$R_2= \left( \dfrac{ v_0^2 }{g} \right)\sin(2\times 41.51)- \left(\dfrac{(2 \times 0.6)v_0^2 }{ (0.5\times9.8)g } \right) \sin^2( 41.51)$$ $$R_2= \left( \dfrac{ v_0^2 }{g} \right)\left[\sin(2\times 41.51)- \left(\dfrac{(2 \times 0.6) }{ (0.5\times9.8) } \right) \sin^2(41.51)\right]$$ $$R_2= \left( \dfrac{ v_0^2 }{g} \right)\left[0.993- 0.108\right]$$ $$R_2= 0.885R_1 \tag 5$$ The reduced percentage of the range is given by $$\dfrac{R_1-R_2}{R_1}\times 100=\dfrac{R_1-0.885R_1}{R_1}\times 100=\color{red}{\bf 11.5\%}$$
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