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

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    Free fall Velocity
AI Thread Summary
The discussion focuses on calculating the ratio of time spent above half the maximum height (t1) to the time taken to reach that height (t2) during a vertical jump. The derived ratio is t1/t2 = 2(sqrt(2) + 1), with some debate on the interpretation of time t = -v0/(sqrt(2)g) as the duration before reaching maximum height. Participants agree that the equation y = 1/2(v0^2/g) - (g/2)t^2 effectively measures time relative to the maximum height. There is a suggestion that a more comprehensive comparison should include all time spent below half height, not just half of it. The final consensus confirms the ratio as accurate, emphasizing the need for clarity in the time comparisons.
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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?
 
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Clockclocle said:
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).
Clockclocle said:
Is it legal to say that t =-(v0)/(sqrt(2)g) is the time before the free fall at ymax happens?
Your equation
Clockclocle said:
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.
 
yes the answer is 2 (sqrt(2)+1). So is it true?
 
Clockclocle said:
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?
 
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