Answer
The total time in the top $15.0~cm$ of this jump is $0.35~s$
Work Step by Step
At the top of the jump, $v = 0$
We can find the time it takes for the player to fall from $76.0~cm$ to $61.0~cm$:
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(0.15~m)}{9.8~m/s^2}}$
$t = 0.175~s$
Since the time to move from $61.0~cm$ up to $76.0~cm$ is also $0.175~s$, the total time in the top $15.0~cm$ of this jump is $0.35~s$
