# Homework Help: Acceleration/Velocity Problem

1. Jan 26, 2008

### ian_durrant

[SOLVED] Acceleration/Velocity Problem

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

Two students, Anne and Joan, are bouncing straight up and down on a trampoline. Anne bounces 2.61 times as high as Joan does. Assuming both are in free-fall, find the ratio of the time Anne spends between bounces to the time Joan spends.

2. Relevant equations

y=V(0)t+ (1/2)gt^2

3. The attempt at a solution

Ok I figured that if Anne is jumping 2.61 times as high as Joan, I could set y (the displacement) as 2.61 for Anne and 1 for Joan. However I plugged the numbers into the equation and ended up getting a negative answer for my time, which doesn't make sense. Here's my equations that I used:

Anne-

2.61=(1/2)(-9.8)t^2
-.53=t^2

Joan-

1=(1/2)(-9.8)t^2
-.204=t^2

I figured what I would do after i got the time is to multiple them both by 2 since I only calcuated them returning for the highest point to lowest point, then plugging them into a ratio. Any thoughts?

2. Jan 26, 2008

### Midy1420

right method, and you won't have to multiple both by 2 since the ratio will cancel that out anyway

3. Jan 26, 2008

### ian_durrant

good point about the ratio, but i'm still confused why i'm going to have a negative number for t, or would this not matter anyways?

4. Jan 26, 2008

### wongdaisiu

Hi...

You might want to think twice about your frame of reference. I work well positive as up and negative is down. But it looks like you work the same way as well since your acceleration is negative. That would mean that your initial location is higher than your final. No matter where you make your y=0...y(initial)>y(final). That is why you are getting a negative.

5. Jan 27, 2008

### ian_durrant

thanks for all the help everyone