Question

If Michael Jordan can jump 48 inches vertically, how much time did jordan spend in the top 6 inches of the jump and how much time in the bottom 6 inches of the jump?

Answer 1

I'll do this in metric units. You can convert later. `48"in"~~1.22m`

If you make an `x-t` diagram of the whole jump, you will see that's a parabola: the first part is the jumping up, decelarating by gravity. At the top his velocity will be zero, and then he falls back.

Let's look at the second half (the falling back).
The equation is `s=1/2g t^2`
Where `g~~9.83=`acceleration of gravity

Fill in what we know:
`s=1.22=0.5*9.83*t^2->t^2=1.22/4.915->t=0.498s`
Now we can calculate the velocity on landing (which is the same velocity at which he jumped up!)
`v=g*t=9.83*0.498=4.90m//s`

So the complete equation for the whole of his jump would be:

`x=v_0 t- 1/2 g t^2=4.90*t-4.915t^2`

You can solve this equation for `x=0.15` The lowest solution gives you half the time he spent below `0.15m` (=on jumping up)

Then solve or `x=1.07`. You get two solutions. Take the difference.
Refer to the graph below to get the picture.
graph{4.9x-4.9x^2 [-0.398, 2.64, -0.11, 1.409]}

(only the part above the `x`-axis is relevant here)