= 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 $(p(1),q(1))=(0,1)$, and
$$
\dot p=0,\qquad\dot q=-\frac{pu+qv}{2\sqrt y}.
$$
Thus $p=0$. Maximizing over $u^2+v^2=1$ gives $u=0,v=1$ as long as $q>0$. The state equation integrates to $\sqrt{y(t)}=1/\sqrt2+t/2$, and the adjoint equation gives $q(t)=\sqrt{y(1)/y(t)}>0$, validating the maximization. Hence
$$
\boxed{u(t)=0,\quad v(t)=1,\quad x(t)=0,\quad y(t)=\left(\frac1{\sqrt2}+\frac t2\right)^2.}
$$
This is globally optimal, not just an extremal: whenever $y>0$, $d\sqrt y/dt=v/2\le1/2$. 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 $\boxed{-y(1)=-(1/\sqrt2+1/2)^2}$.
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