Athelete jump dimensional motion

[madah12]

Homework Statement

In the vertical jump, an athlete starts from a crouch and
jumps upward to reach 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 Y(max) be his maximum height above
the floor. To explain why be seems to bang in the air, calculate the
ratio of the time be is above Y(max/2) to the time it takes him to go
from the floor to that height. You may ignore air resistance.

Homework Equations

v=v₀-gt
delta y = v₀t- 1/2 gt^2

The Attempt at a Solution

so I understood it as it need the ratio of t2 to t1 as t2 the time he need to get from y max/2 to y max and t1 the time from the ground to y max/2
since the displacements are cut two half they are equal so
y₁=v₁t₁-1/2 gt₁^2
=y₂= v₂t₂- 1/2 gt₂^2
where v1 is initial velocity and v2 is velocity at y max/2
v₂=v₁-gt₁
at y max
0=v₂-gt₂
v₂=gt₂
v₁=g(t₁+t₂)
then back to the original equation I can express it with only t1 and t2 as variables
g(t₁+t₂)t₁-1/2 gt₁^2=gt₂^2 - 1/2 g t₂^2

1/2 t₁^2-t₁t₂=1/2 t²₂
I don't know how to work this maybe use to get the ratio should I use the quadratic equation for one of them as a variable?
but I think this is wrong I am sure I took a wrong approach and misunderstood the question I just can't see it.

 

[ehild]

1/2 t₁^2-t₁t₂=1/2 t²₂
I don't know how to work this maybe use to get the ratio should I use the quadratic equation for one of them as a variable?
but I think this is wrong I am sure I took a wrong approach and misunderstood the question I just can't see it.

It is a nice work up to here. Divide the equation by t₂^2, you will get a quadratic equation for the ratio r=t₁/t₂.

ehild

 

[madah12]

I got r= 1 + sqrt(2) is that correct?

 

[ehild]

I just noticed that you have a sign error in the last equation. The previous one is correct yet, you just mistyped the sign of t₁t₂. It should be

1/2 t₁^2+t₁t₂=1/2 t₂^2,

that means r=-1+sqrt(2). It takes a shorter time to reach half of maximum height from ground than the maximum from half-height.

ehild

 

[madah12]

uhm did i switch the times in the original question? was I supposed to get t2/t1 instead of t1/t2?
ratio of the time be is above Y(max/2) to the time it takes him to go
from the floor to that height.
but all I need to do is invert it right so 1/(-1+sqrt(2))=1 + sqrt(2)?

 

[ehild]

Well, yes, I mixed the rtimes... The question was t2/t1, so your result was correct

ehild