University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 61: 2.39


a.) $33.542m \\ $ b.) $15.7\frac{m}{s} \\ $ c.) Blue: acceleration vs time Red: velocity vs time Gray: position vs time

Work Step by Step

Taking the vertical axis up as positive and the surface of mars as $y=0$ Mars gravity: $g_{m} = 0.379g = 0.379\cdot 9.8\frac{m}{s^2} = 3.7142\frac{m}{s^2}$ Time of flight: $t_f = 8,5s$ The tennis ball gets to its maximum height at half of the time of flight. That is $t_{max\ h} = \dfrac{t_f}{2} = \dfrac{8.5s}{2} = 4.25s$ When the tennis ball gets to its maximum height, its velocity will be $0$. That means that $v$ at $t=4.25s$ equals $0$ Therefore we can find how fast was it moving just after it was hit, which is its initial velocity ($v_o$): $$ v_{t=4.25s} = v_o + g_m\cdot t_{max\ h} \\ 0 = v_o - 3.7142\frac{m}{s^2} \cdot 4.25s = v_o - 15.785\frac{m}{s} \\ v_o = 15.785\frac{m}{s} $$ Now we can find how high above its original point did the ball go, that is the maximum height ($y_{max}$): $$ y_{max} = y_o + v_o\cdot t_{max\ h}+0.5\cdot g_m\cdot (t_{max\ h})^2 \\ = 0 + 15.785\cdot4.25s + 0.5\cdot (-3.7142)\cdot (4.25)^2 \\ = 67.086-33.544 = 33.542m $$
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