Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 56: 123

B

Let the positive direction be downward. $1~ft=0.305~m$ $1.62~m/s^2=(1.62~m/s^2)(\frac{1~ft}{0.305~m})=5.31~ft/s^2$ $v_0=0.500~ft/s$, $a=g_{moon}=5.31~ft/s^2$, $\Delta x=4.30~ft$ $v^2=v_0^2+2a\Delta x$ $v^2=(0.500~ft/s)^2+2(5.31~ft/s^2)(4.30~ft)$ $v=\sqrt {(0.500~ft/s)^2+2(5.31~ft/s^2)(4.30~ft)}=6.78~ft/s$