Compare the ratio of two times t1/t2 in this vertical jump

[Clockclocle]

Homework Statement

In the vertical jump, an athlete starts from a crouch and
jumps upward as high as possible. Even the best athletes spend little
more than 1.00 s in the air (their “hang time”). Treat the athlete as a
particle and let ymax be his maximum height above the floor. To explain
why he seems to hang in the air, calculate the ratio of the time he is
above ymax>2 to the time it takes him to go from the floor to that height.
Ignore air resistance.

Relevant Equations

y=(v0)t - 1/2g(t^2)

Here is my attempt. At ymax the velocity turn to zero so we get time t*=v0/g and ymax=1/2 (v0^2/g). At the height y max, since the velocity at this point is 0, i get another equation y= 1/2(v0^2/g)-(g/2)t^2, this equation could be considered as continuation of first equation. Set ymax/2=1/4 (v0^2/g)=1/2(v0^2/g)-(g/2)t^2, I get t=(v0)/(sqrt(2)g) or t =-(v0)/(sqrt(2)g). the negative solution can be thought that the athletes was in ymax/2 v0/(sqrt(2)g second before. so the total time he is above ymax/2 t1==(v0)/(sqrt(2)g)-(-(v0)/(sqrt(2)g))=2(v0)/(sqrt(2)g), and time it take him to ymax/2 is t2=v0/g-(v0)/(sqrt(2)g). So the t1/t2=2sqrt(2)+1. Is it legal to say that t =-(v0)/(sqrt(2)g) is the time before the free fall at ymax happens?

 

[haruspex]

So the t1/t2=2sqrt(2)+1.

You seem to be taking the ratio of the time above half height to the time to reach half height. Is that fair? Also, I think you mean 2 (sqrt(2)+1).

Is it legal to say that t =-(v0)/(sqrt(2)g) is the time before the free fall at ymax happens?

Your equation

y= 1/2(v0^2/g)-(g/2)t^2

effectively defines t as a time measured, forwards or backwards, from the time at which max height is reached.

 

[Clockclocle]

yes the answer is 2 (sqrt(2)+1). So is it true?

 

[haruspex]

yes the answer is 2 (sqrt(2)+1). So is it true?

Yes, but don't you need to compare the time above half height with all of the time below half height, not just half of it?