[SOLVED] Kinematics: Average Deceleration Underwater Hello! First poster here. Here is a tough question from my online HW: 1. The problem statement, all variables and given/known data 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. 2. Relevant equations V^2=Vo^2-2gy V^2=Vo^2+2a(x-xo) Vf=0 3. 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!