1. The problem statement, all variables and given/known data A fire helicopter carries a 564 kg bucket of water at the end of a 19.2 m long cable. Flying back from a fire at a constant speed of 41.6 m/s, the cable makes an angle of 46.0° with respect to the vertical. Determine the force of air resistance on the bucket. Knowns: m = 564 kg d1 = 19.2 m v = 41.6 m/s [tex]\vartheta[/tex] = 46.0 2. Relevant equations For Drag: Fd =(1/2)CdpAv2 3. The attempt at a solution So first I started by making a triangle out of the vertical and findin the values of the sides, the first being d1: d2 = 19.2 sin 46.0 = 13.81 d3 = 19.2 cos 46.0 = 13.38 Then I used the pythagorean theorem (a2 + b2 = c2) to check the values. I also subtracted to find the last angle of the triangle (180-90-46 = 44), but I dont think I needed that. So now I used the drag formula (Fd =(1/2)CdpAv2). I used p =1.29 because thats what it equals at sea level, but since its a helicopter I dont know if I used the right value. How to I calculate this? Then I made Cd 1.15 because that is the drag coefficient of a small cylinder according to my textbook. Is this right? Next, I used the formula A=1/2bh to get A = 92.39 Plugging this in I get Fd = (1/2)(1.15)(1.29)(92.39)(41.62) = 124366.0155 And I know this is wrong. The answer should be 5.72×103 N. Where did I go wrong? And how would I go about fixing this?