Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises - Page 918: 25


$30$ m/s

Work Step by Step

As we know, recall the definition for horizontal distance which says that Horizontal distance during interval $T$ is equal to ( Horizontal speed ) $\cdot$ duration of travel. As we are given that the horizontal speed of a ball is constant. This means that $=v_0 \cos 45^\circ=\dfrac{v_0}{\sqrt 2}$ Need to solve the problem. $90=\dfrac{v_0}{5 \sqrt 2} \times \dfrac{v_0}{\sqrt 2}$ or, $90=(\dfrac{v_0}{5 \sqrt 2}) \times (\dfrac{v_0}{\sqrt 2})$ or, $90=(\dfrac{v_0^2}{10})$ Hence, $v_0=30$ m/s
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