Hello Physics Forum, I am new member and this is my first Thread. While onto the problem. I have been stuck on this particular for well over a week. I do not get the excat answer and to me my solution seems logically. 1. The problem statement, all variables and given/known data An Australian bushman hunts kangaroos with a weapon that consists of a heavy rock tied to one end of a light vine of length 2m. He holds the other end above his head, at a point 2m above the ground and swings the rock in a horizontal circle. the cunning kangaroo has observed that the vine always breaks when the angle measure between the vine and the vertical reaches 60 degrees. At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit? Answer: 3m radius=2m Fc=Tx=tan(60)*mg("According to my free body"), "Vine is 60 degrees from vertical" 2. Relevant equations Centripetal Force Fc=mv^2/r Projectile Motion d=v1t+(1/2)at^2 (vertical) v=d/t(horizontal) 3. The attempt at a solution Centripetal Force Fc=mv^2/r 9.8mtan(60)=(mv^2)/2 9.8tan(60)=v^2/2 v=5.83m/s Projectile Motion This is where I think I am going wrong Vertical Since he is spinning the projectile horizontally, when thrown there will be no vertical velocity v=0 Earlier in the question it said that he was spinning it 2m off the ground d=2 (the rock must travel 2m before landing) a=9.8m/s^2 d=v1t+(1/2)at^2 2=4.9t^2 t=.6389 Horizontal t=.6389 v=5.83m/s d=3.7 m?! Please Help, I have been stuck on this for quite a while..