I'm having problems with the following two problems. Thank you for any help! 72) A skiier coasts down a very smooth 10-meter high slope. If the speed of the skiier on the top of the slope is 5.0m/s, what is his speed at the bottom of the slope? I calculated 2gy= v2 and plugged in v= sq root (2x 9.8m/s2 x10m)= 14m/s Does this look correct? We did an example like this in class but we used vinitial = 0 m/s where its 5.0m/s in this problem so I dont know how to go on from there! 79) In #72, if the skiier has a mass of 60kg and the force of friction retards his motion by doing 2500 J of work, what is his speed at the bottom of the slope? I know that there is an equation v= sq root (vo2 - 2ukmgd) but I dont know where to plug #s in or if this is even the right equation?! Thanks again!