The volleyball is played with a ball of about 270 g and 21 cm in diameter.

Question

In one game, the ball ended up in a pool with so much strength that it almost reached the bottom of it and then went up to the surface.

Calculate:

a) The force exerted by the pool water on the ball.

b) The acceleration with which the ball rose to the surface.

c) Assuming that the density of the balloon was uniform, what part of it stood out from the water?

Answer 1

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/s²). The volume of the ball is 4/3πr³ 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 0.011m³. As the density of water at roughly 20C is 1000kg/m³, the weight of the water displaced and the upward force on the ball is 1000 kg/m³ x 0.011m³ x 9.8 m/s² = 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/s² )/0.270 kg = 389 m/s².

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/m³ x 0.011m³ ) = 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.