Kinematics: Average Deceleration Underwater

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A physics problem involves calculating the average deceleration of a person who jumps from a diving board into water. The individual falls 5.0 meters before entering the water and then decelerates over an additional 1.6 meters underwater. The initial velocity upon hitting the water is calculated to be approximately 9.89 m/s. The correct approach to find deceleration involves rearranging the equation to solve for acceleration, leading to the equation 0 = 98 + 3.2a. After correcting algebraic errors, the solution is confirmed, highlighting the importance of careful calculations in kinematic problems.
RedBarchetta
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[SOLVED] Kinematics: Average Deceleration Underwater

Hello! First poster here. :cool: Here is a tough question from my online HW:

Homework Statement


A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person’s downward motion stops 1.6 m below the surface of the water. Estimate the average deceleration (magnitude) of the person while under the water.

Answer: _blank_ m/s^2.

Homework Equations


V^2=Vo^2-2gy
V^2=Vo^2+2a(x-xo)
Vf=0

The Attempt at a Solution


I asked my physics professor for assistance and tried the problem on my own but came to no avail. I started off by using the first V squared formula and plugging in the known variables to find the velocity of the person as they approach the surface of the water.

Then, I plugged the inital velocity into the 2nd V squared equation, as this is the velocity as the person struck the water, and still couldn't find the correct answer. I must be having an algebraic issue.

Okay, there are two distances: 5 m above the water and 1.6 m below the surface of the water. So there is a combined total of 6.6 meters traveled. For above the water:

Vo or inital velocity is zero.

V^2=Vo^2-2gy
V^2=-2gy
V^2=-2(9.8m/s^2)(5m)
V^2=-98m/s

Then take the absolute value of the the square root?

|V|=Sqroot(98m/s)
V=9.8895 m/s

Now, use the second equation to find out the deceleration under the water:

Xo equals zero.
Vf or final velocity is zero.

V^2=Vo^2+2a(X-Xo)
(Plug in inital velocity when the person struck the water and underwater distance for X.)
V^2=(9.8995m/s)^2+2a(1.6m)
V^2=98 m^2/s^2 + (3.2m)a
(Plug in zero for final velocity and move a to the other side)
a=98 m^2/s^2 + 3.2m

Then...? I am not sure. :confused: Any help would be great. Thank you! :smile:
 
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RedBarchetta said:
V^2=(9.8995m/s)^2+2a(1.6m)
V^2=98 m^2/s^2 + (3.2m)a
(Plug in zero for final velocity and move a to the other side)
a=98 m^2/s^2 + 3.2m

Then...? I am not sure. :confused: Any help would be great. Thank you! :smile:
Welcome aboard! Check your algebra. You have 0 = 98 + 3.2a. You've got to subtract 98 from both sides before you solve for a.
 
Thank you! That worked. I knew I had a fault somewhere in my Algebra. :cool:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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