How High Does a Sphere Rise Above Water After Being Released from the Bottom?

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Homework Statement


A 1kg hollow sphere of volume .00419m^3 is released from rest at the bottom of a 2m deep pool of water (density of water is 1000kg/m^3). The sphere accelerates upwards and flies out of the pool. How high does it get above the pool before coming down. Ignore the brief time when the sphere is only partially submerged and assume there is no air or water resistance.


Homework Equations





The Attempt at a Solution



When I drew the free body diagram, I have Buoyant force pushing up and mg pulling down.

B - mg = ma

I know B = Density_water * Volume displaced * g

I know m, so I solved for a, getting approximately 31.2 m/s^2

However, I don't know what to do now. I know that velocity is zero when it stops going up, but I can't find anything with v = v_0 - gt since I also starts from rest. y = y_0 + v_o*t + (1/2)gt^2 wasn't helpful either.

When I attempted energy conservation,

I did

(1/2)mv^2 + mgh = (1/2)mv_f^2 + mgh_f
(1/2)mv^2 = mgh_f

However, my initial velocity is zero so it's not helpful either.

What should I do?
 
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Break the problem up into two parts. When the sphere is in the water and when it is out. What info would be good to know as it leaves the water? You can use kinematics in this, you just have find an expression that incorporates distance into it. Also, I think you made a math error in your acceleration calculation, it seems too big.
 
Thanks. I will try it again.