Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises - Page 69: 104

Answer

$-4 \sqrt {245}$ or, $-62.61$

Work Step by Step

When the objetc hits the ground at that moment , then $s(a)=0 \implies -4.9a^2+200=0$ After solving we get $a=\sqrt{\dfrac{200}{4.9}}$ ...(1) Now, $v(a)=\lim\limits_{t \to a}\dfrac{s(t)-s(a)}{t-a}=\lim\limits_{t \to a}\dfrac{-4.9t^2+200+4.9a^2-200}{t-a}$ or, $\lim\limits_{t \to a}\dfrac{-4.9t^2+200+4.9a^2-200}{t-a}=-4.9\lim\limits_{t \to a}\dfrac{(t-a)(t+a)}{t-a}=-9.8a$ Fro equation (1), we have $v(a)=-9.8a=(-9.8) (\sqrt{\dfrac{200}{4.9}})=-4 \sqrt {245}$ or, $-62.61$
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