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**[SOLVED] Kinematics: Average Deceleration Underwater**

Hello! First poster here. 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....? Im not sure. Any help would be great. Thank you!

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