Disclaimer
The opinions expressed herein are my own personal opinions and do not represent anyone else's view in any way, including those of my employer.
© Copyright 2009
The opinions expressed herein are my own personal opinions and do not represent anyone else's view in any way, including those of my employer.
© Copyright 2009
(a) How much time passes before the fish’s speed doubles?
(b) How much additional time would be required for the fish’s speed to double again?
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vx = 60 m/s
(a)
Time to get to 120 m/s from horizontal:
120 = sqrt(vx^2 + vy^2)
vy = sqrt(120^2 – 60^2) = 103.92 m/s
t = vy / g = 10.59 s
(b)
Time to get 240 m/s from horizontal:
240 = sqrt(vx^2 + vy^2)
vy = sqrt(240^2 – 60^2) = 232.379 m/s
t = vy / g = 23.688 s
additional time = 23.688-10.59 = 13.1 s
Well… I’m assuming speed is a scalar quantity, not a vector, because negating drag, horizontal velocity should remain constant and vertical velocity is initally zero and so can never mathematically double. So, assuming speed is scalar and not vector (i.e. not dependant on direction):
(a) Speed initial: 60 m/s
Speed doubled: 120 m/s = speed initial + (9.81 m/s^2 * t)
120 = 60 + 9.81t
60 = 9.81t
t = 60 / 9.81
(b) Speed doubled again: 240 m/s
240 = 60 + 9.81t
180 = 9.81t
t = 180/ 9.81