Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.1 - Extreme Values of Functions - Exercises 4.1 - Page 192: 75

Answer

Maximum height $s_0+\frac{v_0^2}{2g}$ meters when $t=\frac{v_0}{g}$ sec.

Work Step by Step

Step 1. Given the function $s=-\frac{1}{2}gt^2+v_0t+s_0$ where $s_0=s(0)$ is the starting height, we can find its derivative as $s'(t)=-gt+v_0$. Step 2. The critical points can be found when $s'=0$ or undefined, which happens only when $t=\frac{v_0}{g}$, which gives $s=-\frac{1}{2}g(\frac{v_0}{g})^2+v_0(\frac{v_0}{g})+s_0=s_0+\frac{v_0^2}{2g}$ and we can identify this as a maximum (it can not be a minimum since $g\gt0$ and $\frac{v_0^2}{2g}\gt0$ ). Step 3. We conclude that the body’s maximum height is $s_0+\frac{v_0^2}{2g}$ meters when $t=\frac{v_0}{g}$ sec.
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