Answer
8.75m/s
Work Step by Step
The range has to be minimum 2.5+1.4=3.9m.
So, $R=\frac{v_{0}^{2}sin\theta}{g}$
or, $v_{0}=\sqrt \frac{gR}{sin2\theta}=\sqrt \frac{9.8\times3.9}{sin(2\times15^{\circ})}=8.75m/s$
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 100: 64
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