Answer
The total time in the bottom $15.0~cm$ of this jump is $~~0.074~s$
Work Step by Step
We can find the initial vertical speed at the start of the jump:
$v^2 = v_0^2+2ay$
$0 = v_0^2+2ay$
$v_0^2 = -2ay$
$v_0 = \sqrt{-2ay}$
$v_0 = \sqrt{-(2)(-9.8~m/s^2)(0.76~m)}$
$v_0 = 3.86~m/s$
We can find the time it takes to jump from the floor up to $15.0~cm$:
$y = v_0~t+\frac{1}{2}at^2$
$0.15 = 3.86~t+4.9~t^2$
$4.9~t^2+3.86~t-0.15 = 0$
We can use the quadratic formula:
$t = \frac{-3.86\pm \sqrt{(3.86)^2-(4)(4.9)(-0.15)}}{2(4.9)}$
$t = \frac{-3.86\pm \sqrt{17.8396}}{9.8}$
$t = -0.825~s, 0.037~s$
Since time is positive, the solution is $t = 0.037~s$
Since the time to fall from $15.0~cm$ down to the floor is also $0.037~s$, the total time in the bottom $15.0~cm$ of this jump is $~~0.074~s$
