Answer
The player spends $~~70.7\%~~$ of the range in the upper half of the jump.
Work Step by Step
From the maximum height $h$, we can find the time it takes to fall to half the maximum height:
$y = \frac{1}{2}a_y~t^2$
$t = \sqrt{\frac{2y}{a_y}}$
$t = \sqrt{\frac{(2)(h/2)}{g}}$
$t = \sqrt{\frac{h}{g}}$
From the maximum height $h$, we can find the time it takes to fall to the ground:
$y = \frac{1}{2}a_y~t^2$
$t = \sqrt{\frac{2y}{a_y}}$
$t = \sqrt{\frac{2h}{g}}$
We can find the fraction of the time spent in the top half of the jump:
$\frac{\sqrt{\frac{h}{g}}}{\sqrt{\frac{2h}{g}}} = \sqrt{\frac{1}{2}} = 0.707$
By symmetry, the fraction of time spent in the upper half of the jump on the way up is equal to the fraction of time spent in the upper half of the jump on the way down.
Since the horizontal range is proportional to the time, the player spends $70.7\%$ of the range in the upper half of the jump.
