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

 

[rwisz]

I would like to commend you on your attempt to solve the problem and your use of the correct formula for projectile motion. However, there are a few things that need to be corrected in order to find the maximum height correctly.

Firstly, the initial vertical velocity (Vyinitial) is not always 0 in projectile motion. In this case, the long jumper's initial velocity is a combination of both horizontal (9.7 m/s) and vertical components. To find the vertical component, we can use the formula V = V0 + at, where V0 is the initial velocity and a is the acceleration due to gravity (-9.8 m/s^2). Since we know that the jumper is in the air for 0.88 seconds, we can plug in these values to find the vertical component of the initial velocity:

V = V0 + at
0 = Vyinitial + (-9.8)(0.88)
Vyinitial = 8.624 m/s

Now, using this value for Vyinitial, we can plug in the values for Yinitial, Vyinitial, t, and a into the formula Y = Yinitial + Vyinitial*t - 0.5*a*t^2 to find the maximum height:

Y = 0 + (8.624)(0.44) - 0.5*(-9.8)(0.44)^2
Y = 1.897 m

Therefore, the maximum height the long jumper reaches is 1.897 m. It is important to note that this is not the same as the height he jumped, which was 8.5 m. The maximum height is the highest point he reaches in the air, while the height he jumps is the distance he covers horizontally before landing. I hope this helps and keep up the good work in your scientific endeavors!