Question 402443
Distance = rate*time only works for constant velocity problems. In this problem, we have the acceleration due to gravity.


The formula for determining position given a constant acceleration is {{{X = (1/2)at^2 + v[0]t + x[0]}}}, where a = -9.81 m/s^2, {{{v[0]}}} is the original velocity, {{{x[0]}}} is the original position (this equation is derived using integral calculus). Here, t = 3 s, {{{v[0]}}} = 30 m/s, {{{x[0]}}} = 160 m. Therefore,


{{{X = (1/2)(-9.81m/s^2)(3s)^2 + (30m/s)(3s) + 160m}}} = 205.9 meters