Answer
$7.20m/s$
Work Step by Step
We can find the required speed as follows:
$\frac{1}{2}mv_A^2+0=0+mgy_B$
This simplifies to:
$v_A=\sqrt{2gy_B}$
We plug in the known values to obtain:
$v_A=\sqrt{2(9.81)(2.64)}$
$v_A=7.20m/s$
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 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 249: 81
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