 Solution

Maximize $y(1)$. Along the positive-$y$ trajectory use Hamiltonian of an optimal-control problem $H=\sqrt{y}(pu+qv)$. The terminal costate is $(p(1),q(1))=(0,1)$, andThus $p=0$. Maximizing over $u^2+v^2=1$ gives $u=0,v=1$ as long as $q>0$. The state equation integrates to $\sqrt{y(t)}=1/\sqrt{2}+t/2$, and the adjoint equation gives $q(t)=\sqrt{y(1)/y(t)}>0$, validating the maximization. HenceThis is globally optimal, not just an extremal: whenever $y>0$, $d\sqrt{y}/dt=v/2\le 1/2$. If a competitor reaches $y=0$ and later becomes positive, apply that bound after its last zero; it has even less remaining time to attain a positive terminal height. A negative terminal height cannot improve this positive maximum. Therefore the minimum objective is $\boxed{-y(1)=-(1/\sqrt{2}+1/2{)}^2}$.