Calculating Ratio of Time Above ymax/2 to Time from Floor to ymax

[Toranc3]

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 he seems to hang in the air, calculate the ratio of the time he is above y_ max/2 to the tme it takes him to go from the floor to that height. You may ignore air resistance.

Homework Equations

y=yo+vo*t+1/2a*t^(2)

The Attempt at a Solution

From floor to ymax

ymax=-g/2*t^(2)

From ymax/2 to ymax

ymax/2 = vo*t - 1/2*g*t^(2)

Not sure what to do after here.

 

[HallsofIvy]

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 he seems to hang in the air, calculate the ratio of the time he is above y_ max/2 to the tme it takes him to go from the floor to that height. You may ignore air resistance.

Homework Equations

y=yo+vo*t+1/2a*t^(2)

The Attempt at a Solution

From floor to ymax

ymax=-g/2*t^(2)

You mean ymax= (g/2)t^2. You don't say how you are setting up your coordianate system, but assuming positive is "up", ymax is positive.

From ymax/2 to ymax

ymax/2 = vo*t - 1/2*g*t^(2)

Please don't use "t" for both times! t in your first equation is the time necessary to get to the highest point. Let's use "s" to mean the time to the half that. Now you have, of course, ymax/2= (g/2)t^2= v_os- (1/2)gs^2. That is a quadratic, (g/2)t^2- v_os+ (g/2)s^2. Solve that for quadratic for s in terms of t, and form the ratio.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Not sure what to do after here. </div> </div> </blockquote>

 

[mfb]

As you have two different times, I would use t and t' to separate them. In addition, try to express vo as function of t or ymax.
You can then use your equations to find an equation with t and t' and no other variables.

 

[Toranc3]

Thanks guys!