The projectile has constant horizontal speed all the time and we can use this formula
x=Vax*t, where t is time it (projectile) takes to reach x=1000m
t=x/Vax=1000/50=20s
As we know we can split projectile trajectory in two parts and the time for each part is equal. We can conclude that time t=20s belong to first part where projectile is going up.
Formula for this type of movement is y=Vay-(gt∧2)/2
If we take acceleration of gravity g=10m/s∧2
we get y=200*20-(10*20∧2)/2=4000-2000=2000m
But if I calculate time it takes the highest point I get
t=Vay/g=200/10=20s
The highest point for that time is
y= H=Vay∧2/2g=200∧2/2*10=40000/20=2000m
This happened because the projecile is fired from the edge of a cliff which is above the sea line.
Good luck!!!