Answer
$d_{1} \approx$ 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
$x=ut+\frac{1}{2}at^{2}$
$d_{1}=0+\frac{1}{2}(9.8)(0.0600)^{2}\approx0.018\; m$
$d_{2}=0+\frac{1}{2}(9.8)(0.12)^{2}\approx0.071\; m$
$d_{3}=0+\frac{1}{2}(9.8)(0.18)^{2}\approx0.16\; m$
