1. The problem statement, all variables and given/known data A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 280 km/h (77.8 m/s). a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3 41a)? Already figured this out and got it right: 538.8 m b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position (Fig. 3 41b)? Magnitude: ??? 2. Relevant equations y=y0+(vy0)t-(1/2)gt^2 3. The attempt at a solution Part A work: y=y0+(vy0)t-(1/2)gt^2 235=(-1/2)(-9.8)(t^2) t=6.93s x=x0+vx0t x=0+(77.8)(6.93) x= 538.8 m ------------------------------------- Part B work d=rt 113.8=77.8t t=1.5 s tfinal= 6.93s - 1.5s tfinal = 5.5s y=y0+(vy0)t-(1/2)gt^2 235=0+(vy0)(5.5)-(1/2)(9.8)(5.5^2) vy0~= 70 m/s I think I did this problem right, but the computer keeps saying that it's wrong. Is this a rounding error? I have tried 70 m/s, but it said that was wrong so I think that I might have to round to 69.7 m/s instead of 70 m/s since the computer is picky sometimes. I only have one chance left to get it right. :P If this isn't a rounding error, what have I done wrong? Any help is appreciated, thanks.