## University Physics with Modern Physics (14th Edition)

$0.0868 m/s^2$.
The rock has a constant downward acceleration in both places. We know the initial velocity of 0 m/s and the displacement. Use the kinematics formulas for constant acceleration to relate the distance fallen to the time. Let the downward direction be positive. The rock has zero initial velocity. Let h be the distance the rock falls. Use $y-y_o=v_{oy}t+0.5at^2$ $h= 0.5(a)t^2$ The rock falls the same distance h in both places, so we have the following: $0.5(a_{Earth})t_{Earth}^2= 0.5(a_{En})t_{En}^2$ $a_{En}=a_{Earth}\frac{ t_{Earth}^2}{ t_{En}^2}$ $a_{En}=9.80m/s^2\frac{(1.75s)^2}{(18.6)^2}$ $a_{En}=0.0868 m/s^2$