# Projectile Motion-long jumper

1. Sep 19, 2010

### atbruick

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Sep 19, 2010

### 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.

3. Sep 20, 2010

### 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.

4. Sep 20, 2010

### 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