In first case, let at point $P$, its kinetic and potential energies are equal ie,
$\frac{1}{2} m v^{2}=m g h$
$\Rightarrow h=\frac{v^{2}}{2 g} \,\,\,\ldots$ (i)
In second case. when body's
velocity is $2 v$ then at the same point $P$
$\frac{ PE }{ KE }=\frac{m g \times \frac{v^{2}}{2 g}}{\frac{1}{2} m(2 v)^{2}}=\frac{1}{4}$