Calculating Maximum Height in Projectile Motion for an Olympic Long Jumper

[atbruick]

Homework Statement

An Olympic long jumper is capable of jumping 8.5 m. How high does he goes? (Assuming he lands standing upright). His initial horizontal velocity is 9.7 m/s.

Before it asked his height, I found that the time he is in the air is 0.88 seconds.

Homework Equations

Y=Yinitial+Vyinitial*t-.5at^2

The Attempt at a Solution

Since Y is the maximum height, I just tried plugging in numbers for the symbols in the rest of the equation. Yintial was 0, Vyintial was 0, t was 0.44 because max height is in the middle, and a is -9.80m/s2. I got an answer of 9.5, which was wrong. I have a feeling I'm substituting incorrectly but can't figure out what.

 

[Delphi51]

The assumption that Vy initial is zero is wrong. If it was zero, then the maximum height would be zero.

It seems to me you would have to use Vf = Vi + at to find Vi before doing your vertical distance calc.

 

[atbruick]

Ok, so hoping my calculations for Vy initial are correct, I got 18 m/s, then I tried finding the maximum height and got 20 meters with time as 0.88, which was incorrect. So I thought I should put in time as 0.44 because that is when he will be at maximum height and got 8.9 and that was incorrect too. Not sure what to do now.

 

[Delphi51]

I think 18 is much too large for Vy initial. It would be more helpful if you showed what you did than what the answer was. I used
Vf = Vi + at
-Vi = Vi -9.8*t
2*Vi = 9.8*0.88