Recent content by Adeopapposaurus

v upstream = 5 - 3 =2 v of the water = 3 arccos = (2/3) = 48.19 v across the river = c^2 - a^2 = b^2 √3^2 - 2^2 = 6m/s t = 80/6 = 13.3 can someone please point out my mistakes? am I even supposed to solve for angle like this or "perpendicular to the shores" just means that the angle is supposed...

ma = mg * sinα - fmg * cosα a = g (sinα-f * cosα) v = g*t(sinα - f * cosα) 14.7 = 10 * 2 (sin60 - f * cos60) f = 0.26 Can someone please check if my solution is correct? I'd really appreciate that.

I know that this should be a very simple problem, but I don't understand how to solve it without knowing the height at which the tank is separated from the rocket. I will be very grateful for any hint.