Answer:
Because of the buoyant force
Explanation:
Jumping into a pool change the amount of apparent gravity acting on a person.
Normally, for a person in free fall, the weight of the person is given by:
[tex]F=mg[/tex]
where m is the mass of the person and g=9.8 m/s^ is the acceleration due to gravity.
When a person is in the water, there is a buoyant force pushing the person upward. The magnitude of the buoyant force is
[tex]B=\rho_w V g[/tex]
where
[tex]\rho_w[/tex] is the density of the water
V is the volume of displaced fluid
g is the acceleration due to gravity
So the net force acting on the person is
[tex]F= mg - \rho_w V g[/tex] (1)
Since V corresponds to the volume of the person, we can rewrite it as
[tex]V=\frac{m}{\rho_p}[/tex]
where [tex]\rho_p[/tex] is the density of the person. Substituting into eq.(1),
[tex]F=mg- m \frac{\rho_w}{\rho_p} g = mg - mg'[/tex] (2)
where we called
[tex]g'=\frac{\rho_w}{\rho_p} g[/tex]
So we can further rewrite (2) as
[tex]F=m(g-g')[/tex]
so we see that the gravity acting on the person, g, has been modified into (g-g') due to the presence of the buoyant force.
