Hi radio,
That is a good question. 
I am not sure if the mass comes into this problem or not. I don't think that it does. 
This is my guess. I will ask someone more knowledgable to take a look at this question though.
I usually do these with calculus.
Initially. When the ball first leaves the ground
$$\\t=0, \\ \quad accel= \ddot y = -10m/s^2, \\ \quad vel=\dot y=u\;\;m/s,\\ \quad displacement=y=0\;metres\\\\
Ongoing\\
\ddot y = -10\qquad $Any stat point will be a maximum$\\
\dot y=-10t+u\\
y=-5t^2+ut\\\\$$
Maximum height will be attained when velocity =0
$$\\-10t+u=0\\
10t=u\\
t=\frac{u}{10}\;\;seconds\\\\
$Now find y when $ t=\frac{u}{10}\\\\
y=-5\times \frac{u^2}{100}+u\times \frac{u}{10}\\\\
y=\frac{-u^2}{20}+ \frac{2u^2}{20}\\\\
y= \frac{u^2}{20}\\$$
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LET me see if I can do the same thing using physics formulas.
Initial velocity =u
final velocity = v =0
accel =a = -10
find displacement s
the only formula with u,v,a and s is [4]
$$\\v^2=u^2+2as\\
0=u^2+2*-10s\\
-u^2=-20s\\\\
s=\frac{u^2}{20}$$
I got the same answer both ways. That is good anyway
The ball will reach a height of $${\frac{{{\mathtt{u}}}^{{\mathtt{2}}}}{{\mathtt{20}}}}$$ metres on the first bounce
