Existence, Uniqueness & Bifurcations · Hard problem

On $t \ge 0$, the IVP $y' = t\sqrt{y}$, $y(0) = 0$ does not have a unique solution. The nonnegative solutions are $y = 0$ for $0 \le t \le c$ and $y = \dfrac{(t^2 - c^2)^2}{16}$ for $t > c$, where $c \ge 0$. If such a solution satisfies $y(3) = 1$, find $c$.