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: 26


$$10 \sqrt{6 g}~m/s\approx 76.68~m/s$$

Work Step by Step

We have the position and velocity functions: \begin{array}{l} s(t)=s_{0}+v_{0} t-\frac{1}{2} g t^{2} \\ v(t)=v_{0}-g t \end{array} then for $s_0=300$ and $v_0=0$, we have $$s(t) =300- -\frac{1}{2} g t^{2} $$ at $s=0$, we get $t=\sqrt{\frac{600}{g}} $ Hence \begin{aligned} v\left(\sqrt{\frac{600}{g}}\right) &= -g \sqrt{\frac{600}{g}} \\ &=\frac{g}{\sqrt{g}} \sqrt{600} \\ &=10 \sqrt{6 g} \end{aligned}
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