Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.12 Chapter Review - Additional Problems - Page 109: 2

Answer

See below

Work Step by Step

Given: $v=80 mph\\ h(0)=2\\ g=32\frac{feet}{sec^2}$ and the distance of the wall is $30$ feet. For the height of the ball: $\frac{dh}{dt}=-gt\\ dh=-gt dt$ Integrate both sides: $\int dh=\int -gt dt\\ h=-g.\int t dt$ The solution is $h(t)=-\frac{1}{2}gt^2+c$ Substitute $h(0)=2 \rightarrow h(0)=-\frac{1}{2}g(0)^2+c\\ \rightarrow c-0=2\\ \rightarrow c=2$ then $h(t)=2-\frac{1}{2}gt^2$ When the ball hits the ground, we have: $h(t)=0\\ \rightarrow 2-\frac{1}{2}gt^2=0\\ \rightarrow \frac{1}{2}gt^2=2\\ \rightarrow gt^2=4\\ \rightarrow t^2=\frac{4}{g}\\ \rightarrow t=\sqrt \frac{4}{g}\\ \rightarrow t=\sqrt \frac{4}{32}=\sqrt \frac{1}{8}=\frac{1}{2\sqrt 2}$ The distance the ball travels before it hits the ground is: $d=v.t\\ d=80.\frac{1}{2\sqrt 2}\\ d=80.\frac{5280}{3600}.\frac{1}{2\sqrt 2}=41.5$ feet We have the length of the court is 40 feet, and since the ball travels 41.5 feet, it will hit the front wall before it falls to the ground.
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