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: 25


$200 ~\mathrm{m} / \mathrm{s}$ $2040.82 \mathrm{~m}$

Work Step by Step

We have the position \begin{align*} s(t)&=s_{0}+v_{0} t-\frac{1}{2} g t^{2}\\ &=200 t-4.9 t^{2}\end{align*} Then the velocity is (derivative of position): $$v(t)=200-9.8 t $$ The maximum velocity of $200 \mathrm{m} / \mathrm{s}$ occurs at $t=0$. The maximum height occurs when $v=0 $ at $t \approx 20.41\mathrm{s}$ Hence, the height is given by $$s(20.41) =200 (20.41)-4.9 (20.41)^{2}\approx 2040.82 \mathrm{~m}$$
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