Answer:
Explanation:
The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.
Change in gravity potential energy = change in spring potential energy
mgh = 1/2kh^2
Assume gravity constant g is 10m/s^2
2.6*10*h = 1/2*200*h^2
100h^2 - 26h = 0
2h(50h - 13) = 0
h = 0 or h = 13/50 = 0.65m
h = 0 is before the spring is stretched
So the maximum distance is 0.65m.
As Potential energy =mgh
m= 0.95kg
h=3 meter
g = 9.8 m/sec^2. ( acceleration due to gravity)
So P.E =(0.95)(9.8)(3)kgm^2/s^2
P.E =27.93 joules
Drag Force = bv^2 = ma; a = g = 9.81 m/s^2
b = mg/v^2 = (0.0023×9.81)/(9.4^2)
b = 0.000255