A diver of mass 60 kg, on entering water with an initial velocity of 16.88 ms^-1

Question

A diver of mass 60 kg, on entering water with an initial velocity of 16.88 ms^-1, experiences a force of resistance, due to the water opposing motion, of magnitude 180 nu Newtons, where nu is the velocity- of the diver in ms^-1. In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720 N. Take the origin of motion to be at the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10 m/s^2. Calculate the maximum depth, in metres, the diver will submerge. Give your answer to 3 decimal places.

Explanation / Answer

according to Newton's second law,

F = m a

mg - F_buoyancy - F_resistance = m a

60(10) - 720 - 180v = 60 (dv/dt)

-120 - 180v = 60 (dv/dt)

all sides divided by 60

dv/dt + 3v = -2

= e^( 3 dt) = e^(3t)

v = - 2 dt + c

v e^(3t) = - 2 e^(3t) dt + c

v e^(3t) = - (2/3)e^(3t) + c

v = -2/3 + c e^(-3t)

initial condition, v = 16.87 at t = 0

16.87 = -2/3 + c e^(-3 * 0)

c = 17.5367

v = -2/3 + 17.5367 e^(-3t)

v = dx/dt

x = v dt

x = (-2/3 + 17.5367 e^(-3t)) dt

x = - (2/3)t - 5.8456 e^(-3t) + C

initial condition, x = 0 at t = 0

0 = - (2/3)*0 - 5.8456 e^(-3*0) + C

C = 5.8456

final solution,

x = - (2/3)t - 5.8456 e^(-3t) + 5.8456

x = - (2/3)t + 5.8456 (-e^(-3t) + 1)


at maximum depth, v = 0

v = -2/3 + 17.5367 e^(-3t)

0 = -2/3 + 17.5367 e^(-3t)

t = 1.0899 s

and maximum depth is

x = - (2/3)t + 5.8456 (-e^(-3t) + 1)

x = - (2/3)(1.0899) + 5.8456 (-e^(-3 * 1.0899) + 1)

x = 4.8968 m

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