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 ball was uniform, what part of it stood out from the water?

Answer 1

Let

m = 270g be the mass of the ball,

r = 10.5cm be the radius of the ball

d = 1 g/cm³ be density of the pool water

g = 9.8 m/s² be gravitational acceleration

V = volume of the ball = (4/3)π r³= (4/3)π 10.5³ = 4849 cm³

a) By the Archimedes' principle, the force F is the difference between the

weight of the water displaced by the ball and the weight of the ball.

F = Vdg - mg =

(Vd - m)g =

(4849 cm³ . 1g/cm³ - 270g) 9.8 m/s² =

44874 gm/s² =

44.894N

b) Assuming no water resistance,

the acceleration comes from Newton's second law

a = F/m = 44.894 N/0.27kg = 166.2 m/s²

c) By the Archimedes' principle, the volume V₀ of the ball that will

remain submerged will be large enough to displace water equal in weight

to the weight of the ball.

V₀dg = mg

V₀ = m/d = 270g/(1g/cm³) = 270cm³

Thus 270 cm³ will be submerged, and the remainder 4579 cm³ will stand out

from the water, which is about 94%.

It is possible that the teacher wants you to notice the relation

between a) and c).

Namely, the volume of the ball that will stand out is (Vd -m)/d.

And notice that the expression (Vd - m) appeared in the calculation of part a).

The reason for this coincidence is that the volume that will stand out

will be what "the water is trying to get rid of", as

it it causing the force pushing the ball out.