Answer
Minimum launch speed to complete the jump $=5.56m/s$
Work Step by Step
$S=ut+\frac{1}{2}at^{2}$
$4=0+4.9t^{2}$
$t=0.9s$
So, required horizontal speed = $\frac{5}{0.9}m/s=5.56m/s$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
Published by
Pearson
ISBN 10:
0-32160-183-1
ISBN 13:
978-0-32160-183-4
Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 99: 54
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