Question 1072666
WITHOUT FORMULAS:
The initial upwards speed was {{{"30 m /s"}}} .
For {{{10s}}} , that upwards speed decreased linearly due to gravity
by {{{"10 m /s"}}} each {{{s}}} {{{"( 10"}}}{{{m}}}{{{"/"}}}{{{s^2}}}{{{")"}}} ,
for a total decrease of {{{10*10}}}{{{"m /s" = 100}}}{{{"m /s"}}}
to {{{"30 m /s"-"100 m /s"="- 70 m /s"}}} .
The average upwards speed for those {{{10s}}} was
{{{(30+(-70))/2}}}{{{"m /s" = -20}}}{{{"m /s"}}} .
At an average speed of {{{-20}}}{{{"m /s"}}} for {{{10s}}} ,
the stone's upwards displacement was
{{{(10s)}}}{{{"(-20 m / s ) = -200 m"}}}
So, the stone left Peter's hand {{{highlight(200m)}}} above the surface of the water.
 
WITH FORMULAS:
You may have learned in physics class that, for an object moving linearly under constant acceleration,
its position {{{(x)}}} as a function of time {{{(t)}}} can be modeled as
{{{x(t)=x[0]+v[0]t+(1/2)at^2}}} , where
{{{a}}} is the constant acceleration, while 
{{{x[0]}}} and {{{v[0]}}} are respectively the position and velocity of the object at {{{t=0}}} .
You could choose to measure position as height (in meters) above the water surface,
and time (in seconds) since the stone left Peter's hand.
In that case, distance above the water, upwards velocity and upwards acceleration are positive,
so initial velocity is {{{v[0]=30}}}{{{"( m / s )"}}} ,
acceleration is {{{a=-10}}}{{{"( m /"}}}{{{s^2}}}{{{")"}}} ,
{{{x[0]}}} is what you are looking for ,
and you know that {{{x(10)=0}}} at {{{t=10}}} .
Substituting into {{{x(t)+x[0]+v[0]t+(1/2)at^2}}} ,
{{{x(10)=x[0]+30*10+(1/2)(-10)*10^2}}} ,
{{{0=x[0]+300-5*100}}} ,
{{{0=x[0]-200}}} , and {{{x[0]=highlight(200)}}} (in meters , of course).