A detective of 50 Kg is released at rest from the deck of an aircraft career int

Question

A detective of 50 Kg is released at rest from the deck of an aircraft career into the ocean and allow it to sink. Assume that the deck is 19.6 meter above the water surface and we can neglect the air resistance. While gravity (g=9.8 m/s^2) is pulling the device down, a buoyancy force of 1/5 times the weight of the device is pushing it up. If we assume that the water resistance exerts a force on the device that is proportional to the velocity of the device and the proportional constant is 10n-sec/m, find the speed of the device and its position t second after it touches the water surface.

Explanation / Answer

the total forces on the device are:

mg-Fbuyoncy-Fresistance=ma=mv'

Fb=0.2m=0.2*50=10[N]

Fr=10v

so we get:

50*9.8-10-10v=50v'

v'=9.6-0.2v

dv/dt=9.6-0.2v

dv/(9.6-0.2v)=dt

ln(9.6-0.2v)/-0.2=t+C1

ln(9.6-0.2v)=-0.2t+C2

9.6-0.2v=e-0.2t+C2=C3e-0.2t

v=Ce-0.2t-48

let's find the initial conditions:

x=x0+v0t+0.5at2

0=19.6-0.5*9.81*t2 --> t3[sec]

v=v0+at=0-9.8*3=-29.4[m/sec]

If we use it in our problem we'll get:

v(0)=29.4 (in ths case it's positive since we've used the downward direction as positive, and in the velocity analysis we've used the upward direction as positive)

so we get v(0)=C-48=29.4 --> C=77.4[m/sec]

v=77.4e-t/5-48

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