Homework Help: Athelete jump dimensional motion

1. Sep 14, 2010

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
v=v0-gt
delta y = v0t- 1/2 gt^2

3. 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 dont 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.

2. Sep 14, 2010

ehild

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

3. Sep 14, 2010

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

4. Sep 14, 2010

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

5. Sep 14, 2010

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)?

6. Sep 14, 2010

ehild

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

ehild