Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 45

$d_{1} \approx$ 0.018 m $d_{2} \approx$ 0.071 m $d_{3} \approx$ 0.16 m

Work Step by Step

Let down be the positive direction $x=ut+\frac{1}{2}at^{2}$ $d_{1}=0+\frac{1}{2}(9.8)(0.0600)^{2}\approx0.018\; m$ $d_{2}=0+\frac{1}{2}(9.8)(0.12)^{2}\approx0.071\; m$ $d_{3}=0+\frac{1}{2}(9.8)(0.18)^{2}\approx0.16\; m$

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