Long Jump problem with only distance and height given

[Paulo Serrano]

Homework Statement

[PLAIN]http://img199.imageshack.us/img199/2586/physicshelp.jpg

1. What is the time interval between the initial jump and when the center of mass reaches the maximum height?
2. Average horizontal velocity during the jump.
3. The time interval between the time when the center of mass was at the maximum height and the end of the jump.

Homework Equations

I assume this would work as a projectile motion question, but I can't think of how to do it without having at least the angle or initial velocity.

The Attempt at a Solution

I really don't know where to start. This isn't homework, it's a sample question for the specialized university entrance exam. (University of Sao Paulo Vestibular)

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I'm hoping this is easier than it looks and I'm just being dense. Thanks in advanced to whoever can give me a hand studying. I still have until November to get a grasp on this stuff and until January to actually master it.

EDIT: I guess it is much easier than I though. So far I know the initial velocity and the time. That's question number one answered. I have to go now though so I'll see if I can finish later.

 

[jhstroud]

Start by thinking about it as a conservation of energy problem, where the change in potential energy equals a change in kinetic energy. That's how I solved it.

 

[Redbelly98]

I don't think conservation of energy will work here, since it is asking for the time interval.

Paulo, are you familiar with kinematic equations for constant acceleration? They should be in any physics textbook that covers this material.

 

[Paulo Serrano]

Yeah, Redbelly, I am. I did the following:

Found the initial velocity in the vertical direction using V_f^2-V_i^2 = 2∙a∙∆d and then used ∆V=at to find the time from jump to max height.

After that I had to calculate the time it took the jumper to travel the excess 1.04m on the y-axis using plain old ∆d=V∙t. I added that time interval to the time interval it took to the first time to get question 3.

I ended up trying to get all the information possible out of the image so I could check my work. http://www65.wolframalpha.com/input...e.v_7.73+m/s&f4=39.8&f=Projectile.alpha_39.8", I was able to ensure I'm at least partially right. Not sure about the final time interval, but I'm pretty confident.

 

[Redbelly98]

Yeah, Redbelly, I am. I did the following:

Found the initial velocity in the vertical direction using V_f^2-V_i^2 = 2∙a∙∆d and then used ∆V=at to find the time from jump to max height.

Looks good so far.

After that I had to calculate the time it took the jumper to travel the excess 1.04m on the y-axis using plain old ∆d=V∙t. I added that time interval to the time interval it took to the first time to get question 3.

The figure doesn't show the 1.04m, I guess that information is somewhere in the problem statement.
The problem here is that it is still accelerated motion, so ∆d=V∙t won't work. But you can use the same procedure as in (1) to find ∆V first.

 

[Paulo Serrano]

The 1.04m is what's left after the "regular jump". In normal conditions, wouldn't the first half of the jump be the same as the second half? Since the jumper is jumping from a slightly elevated terrain and her center of mass is much lower upon landing, it has more time to fall, and consequently more time to travel horizontally.

I got the 1.04m by subtracting 6 (the distance it would be if it was a normal jump) from 7.04m.

 

[Redbelly98]

I see, you're right. 1.04m is the extra horizontal distance at the end of the jump. I thought you meant a vertical distance, which is why I was confused.

As you're probably aware, you need to use the horizontal velocity in ∆d=V∙t

 

[Paulo Serrano]

Yep. It actually ended up being much simpler than it looked. When I first looked at and saw that neither the angle or initial velocity was given I thought it was going to be more complicated. I think I learned my lesson: don't give up before actually trying.