Answer
$1.40\ m$
Work Step by Step
Forces acting on the ball are
Upward: buoyant force ... $F_{b}=\rho_{water}\cdot V_{ball}\cdot g$
Downward: weight of the ball: $mg=\rho_{ball}\cdot V_{ball}\cdot g$
Net force exists, accelerating the ball with acceleration $a:$
$\rho_{water}\cdot V_{ball}\cdot g-\rho_{ball}\cdot V_{ball}\cdot g=\rho_{ball}\cdot V_{ball}\cdot a$
... We solve this for $a.$
$a=\displaystyle \frac{gV_{ball}(\rho_{water}-\rho_{ball})}{\rho_{ball}\cdot V_{ball}}=\frac{g(\rho_{water}-\rho_{ball})}{\rho_{ball}}=g(\frac{\rho_{water}}{\rho_{ball}}-1)$
... given that $\displaystyle \frac{\rho_{ball}}{\rho_{water}}=0.300,$
$a=(9.80m/s^{2})(\displaystyle \frac{1}{0.300}-1)=22.9m/s^{2}$
The ball accelerates upwards for a displacement of $x-x_{0}=0.600m$.
Use formula 2-16 from Table 2-1 (for constant acceleration motion).
The velocity at the surface is given by
$v^{2}=v_{0}^{2}+2a(x-x_{o})$
$v=\sqrt{(22.9m/s^{2})(0.600m)}=5.24m/s$
Assuming no reduction of velocity due to energy transfer (splashing, waves, etc),
this is the initial velocity of a vertical projectile motion of the ball, assuming no air resistance.
We want the displacement $(x-x_{0})$ when $v_{0}=5.24m/s$ and $a=-g$,
and the final velocity is zero (at the topmost point in the trajectory)
$v^{2}=v_{0}^{2}-2g(x-x_{o})$
$0=(5.24m/s)^{2}-2(9.81m/s^{2}) (x-x_{0})$
$x-x_{0}=\displaystyle \frac{(5.24m/s)^{2}}{2(9.81m/s^{2})}=1.40m$
