At a quarry where the bottom is lled with water known to be 3 meters deep, an ob

Question

At a quarry where the bottom is lled with water known to be 3 meters deep, an object with a mass of 3 kg is dropped o of a sheer edge. It falls for 2.5 seconds, hits the water, and sinks to the bottom of the water in a path that is more or less a straight line down. It is released from rest.
From previous experiments, we estimate the terminal velocity of similar objects to be 250 m/sec in air, and 4 m/sec in water.
Furthermore, we'll assume that the drag force in air is numerically equal to some constant times the velocity; but the drag force in water is numerically equal to some constant times the square of the velocity. (The power on the velocity comes from estimates on how much turbulence is created by the object as it moves through a medium; the more turbulence, the more accurate the V^(2) model is.)
So we have two initial value problems: If our coordinate system is oriented so that down is the positive direction, we have
m((dV)/(dt))=mg - K1*V V(0)=0
m((dV)/(dt))=mg - K2 * V^(2) V(2.5)=Vs
We don't know what vs (s for ''splashdown") is for now; we also don't know K1 or K2 but it may be possible to determine these by thinking about the terminal velocities.
I would like you to provide the following:
1. A method of finding K1 and K2

Explanation / Answer

finding K1 and K2 can be done using the terminal velocities which means that the velocity can't exceed that meaning that accelaration((dV)/(dt)) at that point=0

from eqn m((dV)/(dt))=mg - K1*V we get

mg-K1V=0

or K1=mg/V=3*9.8/250=0.1176kg/sec

similarly from eqn m((dV)/(dt))=mg - K2 * V2 we get

mg - K2 * V2=0

or K2=mg/V2=3*9.8/4*4=1.8375kg/m

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