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 - Review Exercises - Page 91: 38

Answer

$$v=-70$$

Work Step by Step

The object will impact the ground when $s(t)$ becomes zero; that is,$$s(t)=-4.9t^2+250=0 \quad \Rightarrow \quad t=\frac{50}{7}$$.Now, we can find the velocity of the object when it impacts the ground by substituting $a=\frac{50}{7}$ into the definition of velocity given in the question as follows.$$v(a=\frac{50}{7})=\lim_{t \to \frac{50}{7}}\frac{s(\frac{50}{7})-s(t)}{\frac{50}{7}-t}=\lim_{t \to \frac{50}{7}}\frac{(-4.9((\frac{50}{7})^2)+250)-(-4.9t^2+250)}{\frac{50}{7}-t}=\lim_{t \to \frac{50}{7}}\frac{4.9(t^2-(\frac{50}{7})^2)}{\frac{50}{7}-t}=\lim_{x \to \frac{50}{7}} - \frac{4.9(t+\frac{50}{7})(t-\frac{50}{7})}{t-\frac{50}{7}}=\lim_{x \to \frac{50}{7}}-4.9(t+\frac{50}{7})=-70.$$
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