Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.4 Rates of Change - Exercises - Page 130: 27


$$19.6 \mathrm{m}$$

Work Step by Step

We have the position function \begin{align*} s(t)&=s_{0}+v_{0} t-\frac{1}{2} g t^{2}\\ &=v_{0} t-4.9 t^{2}\end{align*} When the ball hits the ground after $4$ seconds, its height is $0$: \begin{align*} 0&=s(4)\\ &=4 v_{0}-4.9(4)^{2} \end{align*} then $v_{0}=19.6 \mathrm{m} / \mathrm{s} .$ The ball reaches its maximum height when $v=0,$ \begin{align*} v&= v_0-9.8t\\ 0&=19.6-9.8t\\ \end{align*} Hence, $t=2$ At $t=2 $, , we have the height \begin{align*} s(2)&=0+19.6(2)-\frac{1}{2}(9.8)(4)\\ &=19.6 \mathrm{m} \end{align*}
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