1. An object of unit mass is dropped from a height of 20 above a liquid-filled reservoir of depth 50. If the acceleration of gravity is 9.8 and the resistance-to-motion coefficients of air and liquid are 1 and 4, respectively, compute the time taken by the object to hit the bottom of the reservoir. 2. ma=mg-rv , or v'+(r/m)v=g , also the book gives an answer of t= approx. 22.7, but I can't figure out how to get to this answer. 3. This is a linear equation, so I know I need to use an integrating factor to solve. From an example in the book, after finding the integrating factor they obtain solutions of v(t)=(mg/r)+(v0-(mg/r))e^-(r/m)t, t>0 , and also y(t)=(mg/r)t+(m/r)((mg/r)-v0)((e^-(r/m)t)-1)+y0, t>0. I thought I should split this into two problems first. I tried plugging in the numbers to find the time t when the object reaches the origin (the water) by setting v(t)=0, thinking that I could then plug this time into y(t). The thing is though, I don't think this is correct because the object doesn't really hit zero velocity when it hits the water as it would if it was just hitting the ground. I'm really not sure how to precede from here. Any help or suggestions are greatly appreciated, thanks.