(a) 0.40 s
First of all, let's find the initial speed at which Jordan jumps from the ground.
The maximum height is h = 1.35 m. We can use the following equation:
![v^2-u^2=2gh](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2gh)
where
v = 0 is the velocity at the maximum height
u is the initial velocity
is the acceleration of gravity
Solving for u,
![u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s](https://tex.z-dn.net/?f=u%3D%5Csqrt%7B-2gh%7D%3D%5Csqrt%7B-2%28-9.8%29%281.35%29%7D%3D5.14%20m%2Fs)
The time needed to reach the maximum height can now be found by using the equation
![v=u+gt](https://tex.z-dn.net/?f=v%3Du%2Bgt)
Solving for t,
![t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv-u%7D%7Bg%7D%3D%5Cfrac%7B0-5.14%7D%7B-9.8%7D%3D0.52s)
Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of
h' = 1.35 - 0.20 = 1.15 m
Using again the equation
![v'^2-u^2=2gh'](https://tex.z-dn.net/?f=v%27%5E2-u%5E2%3D2gh%27)
we find
![v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s](https://tex.z-dn.net/?f=v%27%3D%5Csqrt%7Bu%5E2%2B2gh%7D%3D%5Csqrt%7B5.14%5E2%2B2%28-9.8%29%281.15%29%7D%3D1.97%20m%2Fs)
And the corresponding time is
![t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7Bv%27-u%7D%7Bg%7D%3D%5Cfrac%7B1.97-5.14%7D%7B-9.8%7D%3D0.32s)
So the time to go from h' to h is
![\Delta t = t-t'=0.52-0.32=0.20 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20t-t%27%3D0.52-0.32%3D0.20%20s)
And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:
![\Delta t=2\cdot 0.20 = 0.40 s](https://tex.z-dn.net/?f=%5CDelta%20t%3D2%5Ccdot%200.20%20%3D%200.40%20s)
(b) 0.08 s
This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:
![v'^2-u^2=2gh'](https://tex.z-dn.net/?f=v%27%5E2-u%5E2%3D2gh%27)
![v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s](https://tex.z-dn.net/?f=v%27%3D%5Csqrt%7Bu%5E2%2B2gh%7D%3D%5Csqrt%7B5.14%5E2%2B2%28-9.8%29%280.20%29%7D%3D4.74%20m%2Fs)
And the corresponding time is
![t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7Bv%27-u%7D%7Bg%7D%3D%5Cfrac%7B4.74-5.14%7D%7B-9.8%7D%3D0.04s)
So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:
![\Delta t = 2\cdot 0.04 =0.08 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%202%5Ccdot%200.04%20%3D0.08%20s)