An eagle is flying horizontally at 7.3 m/s with a fish in its claws. It accidentally drops the fish.(a) How much time passes before the fish's speed quadruples?2.88 s (i got this right)(b) How much additional time would be required for the fish's speed to quadruple again?_________ s
doubling the zero vertical velocity does not make sense because the vectored velocity is what they are talking about. Another thing I find wrong in his work is using 10m^2/sec^2 for G. I would substitute 9. 8 m^2/sec^2. Then only report a couple of figures. 14.6^2 = 7^3^2 + vy^2 159.87 = vy^2 12.644= vy 12.644= G t 1.3 = t (two sig figs) 29.2^2=7.3^2 + vy^2 852.64 - 53. 29 = vy^2 799.35= vy^2 28.28 = vy 28.28 = G t 2.9 = t (two sig figs again) 2.9-1.3 = 1. 6 sec additional time Morningf is substituting vertical velocity for vectored velocity in the second part. The vertical component would be sqrt 15 not 17