# Time above ground question involving ratio

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 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 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.

v = u + at
s = ut + 1/2at^2

The wording of this question is really confusing me. So I need to calculate the ratio of the time he is above ymax/2 (t1) : the time it takes him to go from the floor to ymax/2 (t2)? So its just the ratio t1:t2?

mfb
Mentor
Right.
While it is not necessary, I think it is useful to calculate both t1 and t2, and calculate the ratio based on those two values.

Right.
While it is not necessary, I think it is useful to calculate both t1 and t2, and calculate the ratio based on those two values.

Ok I think I've calculated t1, but I don't think it is correct.. if you are doing a calculation for an object being thrown in the air, until it reaches the point where it is coming back down can you set v=0?

mfb
Mentor
Ok I think I've calculated t1, but I don't think it is correct..
It would be interesting to see your work.
if you are doing a calculation for an object being thrown in the air, until it reaches the point where it is coming back down can you set v=0?
At both points, the object is moving...

If moving up and down together takes 1 second, how long does it take to go up?

It would be interesting to see your work.
At both points, the object is moving...

If moving up and down together takes 1 second, how long does it take to go up?

ok so here is what I have done:

s = vt - 1/2at^2 -- I have never used this formula before, found it off google

s = ymax
a= -9.81 m/s^2
t = 0.5 (because that is half the total time to go up and down and I am just finding the up)
v = 0

ymax = 9.81 x 1/2 x 0.5^2 = 1.23m, so ymax/2 = 0.61m

mfb
Mentor
That is a good approach. Now you just need the time from the ground to 0.61m, or from 0.61m to 1.23m. Alternatively, use the symmetry and calculate the time from 1.23m to 0.61m.