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 =
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.