An 80kg anchor is dropped in the water and plummets to the ocean floor with an observed acceleration of 3.0m/s

Question

what is the magnitude of the water resistance acting on the anchor ? How fast is the anchor traveling after 3.0 seconds? an How long does it take the anchor to drop 30m?

Answer 1

3.0m/s is not an acceleration, it is a speed.
Is the acceleration constant?

Answer 2

I will assume that you really mean 3.0m/s^2 not 3.0m/s.

Therefore there is a net force acting on the anchor. This net force is equivalent to its weight minus the water resistance. The net force is also equal to ma, from the formula: F=ma, where F is the net force, m the mass, and a the acceleration. Thus

Net force,F=weight of anchor-water resistance
Let's call the water resistance, BF.

F=80g-BF
and
F=80*3

80g-BF=80*3
BF=80g-80*3
=80*9.8-80*3
=544N

Let's assume an initial velocity of zero when it was dropped. To solve for the final velocity, v, after 3 seconds, use the formula:

v=u+at
v=0+3*3
=9m/s

To solve for the time it takes to drop 30m, use the formula:

s=ut+1/2at^2, where s is the distance of 30m, u the initial velocity of zero, a the acceleration, and t is the time. Now substitute known values:

30=0*t+1/2*3t^2
30=1.5t^2
t^2=30/1.5
=20
t=4.47s

Answer 3

I am taking Physics 1 next semester although I am not looking forward to it.