Anthony T. answered • 10/29/20

Patient Math & Science Tutor

The force by the water on the object is equal to the weight of the water displaced by the object. To determine this, the volume of the ball and the density of the water is required. The weight of the displaced water is given by d_{h20 }x V_{ball} x g where g is the gravitational acceleration (9.8 m/s^{2}). The volume of the ball is 4/3πr^{3} where r is the radius of the ball, 10.5 cm. Changing cm to m, the radius becomes 0.138m. The volume of the ball is therefore 0.011m^{3}. As the density of water at roughly 20C is 1000kg/m^{3}, the weight of the water displaced and the upward force on the ball is 1000 kg/m^{3} x 0.011m^{3} x 9.8 m/s^{2} = 107.8 N.

The acceleration of the ball while returning to the surface is given by dividing the net upward

force by the mass of the ball. The equation is acc = (F_{buoyant - }mass_{ball} x g)/mass_{ball }where g is the acceleration of gravity. The result is (107.8 N - 0.270 kg x 9.8m/s^{2 })/0.270 kg = 389 m/s^{2}.

To find the fraction of the ball's volume in the water while floating, equate the buoyant force on the submerged volume to the weight of the ball, d_{h20} x p x V_{ball} x g = mass_{ball} x g. Solve for p which is the fraction of the ball's volume under water to get p = mass_{ball} /(d_{h20} x V_{ball}). Numerically this is p = 0.270kg/(1000kg/m^{3} x 0.011m^{3} ) = 0.0245. The fraction of the ball out of the water would be 1.000 - 0.0245 = 0.976.

These numbers seem strange. Please check my math.

Anthony T.

another correction, the upward acceleration is 1714 m/s^2 not the upward force.10/30/20

Anthony T.

I checked my own math and found a mistake. The diameter of the ball in meters is 0.105 m not 0.138m. The actual volume of the ball is 0.00485m^3. The buoyant force is 47.5N. The upward force is 1714 m/s^2. The fraction of the ball's volume out of water is 0.94. I apologize for the error.10/30/20