Why Does an Athlete Seem to Hang in the Air During a Vertical Jump?

[Amar.alchemy]

Homework Statement

Challenge Problem(2.96) from University Physics textbook:
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 1.00s 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 y/2 to the time it takes him to go from the floor to that height. You may ignore air resistance.

Homework Equations

constant acceleration equations: y=Y₀ + V₀t - 4.9t²...1
V²_y=V²_0y-2g(y-y₀)...2

The Attempt at a Solution

Part 1(Time it takes him to go from floor to Ymax /2):
a_y= -g, origin at the floor, V_0y=0, y=Ymax /2, y₀=0

so if i substitute these known quantities in the second equation, then for the velocity at the position Ymax /2 i am getting complex roots. Kindly inform me where i am going wrong.

Thanks,

 

[womfalcs3]

Try using a positive value for g. Since when you ultimately calculate time, using a negative value would result in a negative value under the square root.

Edit: Strike out what I said.

 

[Cyosis]

No. if v0y=0 then how can he even get into the air? You can calculate what v0y is because you know what vymax is. What is vymax?

 

[Amar.alchemy]

No. if v0y=0 then how can he even get into the air? You can calculate what v0y is because you know what vymax is. What is vymax?

Thanks Cyosis, I got it. since we know that Vymax is zero we can calculate the initial velocity of the part 2 which becomes the final velocity of part 1. am i right??

 

[Cyosis]

Yep.