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Drew a staight line down in quadrant two and made a right triangle. Then: tan(62.5°) multiplied by 1.25 mi/h = 2.40 mi/h.

Then figure the time out by distance of 0.505 mi divided by 2.40 mi/h = .210 h.

Then figure upstream at the opposite shore by: (0.505 mi) x (.210 h) = .106 mi = 560.0 feet.

I'm confused because I have a vector pointing east and one pointing north at 62.5° north of west and I feel like I'm ignoring the east pointing vector and that I need to do something else. I don't understand WHY though.