Answer
The ball was $1.15\text{ s}$ in the air.
Work Step by Step
We know that
$t=\frac{2(v_{\circ}sin \theta)}{g}$
We plug in the known values to obtain:
$t=\frac{2(9.85m/s)sin35.0^{\circ}}{9.8m/s^2}$
$t=1.15s$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 106: 32
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