Answer
$5.9\times10^3~m$
Work Step by Step
$v=340~m/s$, $v_0=0$, $a=g=9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$(340~m/s)^2=0^2+2(9.81~m/s^2)\Delta x$
$\Delta x=\frac{(340~m/s)^2}{2(9.81~m/s^2)}=5.9\times10^3~m$
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 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 54: 98
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