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, [tex](g/2)t^2- v_os+ (g/2)s^2. Solve that for quadratic for s in terms of t, and form the ratio.

Not sure what to do after here.

 

[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!

 

[asteriatic]

I would first clarify with the instructor or the source of the problem what exactly is meant by "hang time." Is it the total time the athlete is in the air, or the time from the moment they reach their maximum height to the moment they land back on the ground? This will affect the calculation of the ratio.

Assuming "hang time" refers to the total time in the air, the ratio of time above ymax/2 to time from floor to ymax can be calculated as follows:

Time above ymax/2 = t - (t/2) = t/2
Time from floor to ymax = t/2

Therefore, the ratio is 1:1 or simply 1.

This means that the athlete spends the same amount of time above ymax/2 as they do from the floor to ymax. This can be explained by the fact that the athlete experiences the same acceleration due to gravity during both of these intervals, causing them to reach their maximum height and descend at the same rate.

If "hang time" refers to the time from reaching maximum height to landing, then the ratio would be 1:2, as the athlete spends twice as much time in the air after reaching their maximum height compared to the time it takes to reach that height from the floor.