Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 717: 74

Answer

a) $D=\frac{H(1+r)}{1-r}$ b) $\sqrt (\frac{2H}{g}) \times \frac{1+\sqrt r}{1-\sqrt r}$ c) $T=\sqrt{\frac{2H}{g}} \times \frac{1+k}{1-k}$

Work Step by Step

a) The ball falls an initial height of $H$ meters then climbs a height of $rH$ meters. Then falls to a height of $rH$ meters and climbs a height of $r^{2}H$ meters. Then falls to a height of $r^{2}H$ meters and then climbs a height of $r^{3}H$ meters. Then falls to a height of $r^{3}H$ meters and so on... $D=H+rH + rH +r^{2}H +r^{2}H +r^{3}H +r^{3}H+...$ $D=H + 2rH +2r^{2}H +2r^{3}H...$ $D=-H + 2H +2rH +2r^{2}H+2r^{3}H...$ $D= [\Sigma^{\infty}_{n=0}2Hr^{n}] - H$ $D=[2H\Sigma^{\infty}_{n=0}r^{n}] - H$ $D=[2H\frac{1}{1-r}] - H$ $D=\frac{2H}{1-r}-H$ $D=\frac{2H-H(1-r)}{1-r}$ $D=\frac{2H-H+Hr}{1-r}$ $D=\frac{H+Hr}{1-r}$ $D=\frac{H(1+r)}{1-r}$ b) $\frac{1}{2}gt^{2}_{n}=Hr^{n}$ $t^{2}_{n}=\frac{2H}{g}r^{n}$ $t_{n} = \sqrt (\frac{2H}{g})(\sqrt r)^{n}$ $T=t_{0} +2t_{1} +t^{2}_{2}+...=\Sigma^{\infty}_{n=0}2\sqrt (\frac{2H}{g})(\sqrt r)^{n}-t_{0}$ $=2\sqrt (\frac{2H}{g})\Sigma^{\infty}_{n=0}2(\sqrt r)^{n}-\sqrt (\frac{2H}{g})$ $=\sqrt (\frac{2H}{g})(\frac{2}{1-\sqrt r}-1)$ $=\sqrt (\frac{2H}{g}) \times \frac{1+\sqrt r}{1-\sqrt r}$ c) Calculate the time it takes to decelerate down from $k^{n}v$ meters per second to $0$, or equivalently accelerate from $0$ to $k^{n}v$. $gt_{n}=k^{n}v$ $t_{n} = \frac{k^{n}v}{g}$ $T= t_{0} +2t_{1} +2t_{2} +.... = \Sigma^{\infty}_{n=0} \frac{2v}{g}k^{n} - \frac{v}{g}$ $=\frac{v}{g}(\Sigma^{\infty}_{n=0}2k^{n}-1)$ $=\frac{v}{g}(\frac{2}{1-k}-1)$ $=\frac{v}{g} \times \frac{1+k}{1-k}$ We also know from physics that: $v_f^2=v_i^2+2gH$ Since the ball is dropped with initial velocity 0, $v_i=0$: $v_f^2=0+2gH$ $v=\sqrt{2gH}$ Therefore: $T=\frac{\sqrt{2gH}}{g} \times \frac{1+k}{1-k}$ $T=\sqrt{\frac{2H}{g}} \times \frac{1+k}{1-k}$
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