- #1

DeltaIceman

- 14

- 0

## Homework Statement

A diver springs upward from a board that is 3.05 m above the water. At the instant she contacts the water her speed is 8.51 m/s and her body makes an angle of 75.1 degrees with respect to the horizontal surface of the water. Determine her initial speed as she leaves the board.

## Homework Equations

The equations I have to use for this problem were

(Vf^2)=(Vo^2)+2a(x)

x=1/2(Vo+Vf)t

x=(Vo)(t)+1/2(a)(t^2)

Vx=Vo+at

******Note all these equations are interchangable and can be used to find the y components also********

Basic trig functions are also allowed to be used such as cos=x/r and rearranging it to find the values you need. These trig functions also include the pathagorian theroem (sp?)

## The Attempt at a Solution

This is all my work that I've done for the equation or at least my last attempt. I tried it a couple different ways but this is the only one I felt like I was getting anywhere. Also The equation I used to get Vo or initial velocity was (Vf^2)=(Vo^2)+2a(x).

I'm sure I'm making a dumb mistake. I went through the rest of the homework problems fairly easy. Then I got to this one and was dumbfounded and decided to seek help elsewhere. I wasn't exactly sure if I could assume the height was 3.05m but I started thinking about the problem and since it states that we want the initial speed as she leaves the board. We could consider that velocity to be almost instantaneous and since the elapsed time is so small that the height was basically 3.05 m. But then again its magnitude is going upward and I wasn't exactly sure how that would through off the height. Anyway If you guys can help me it would be greatly appreciated. I am sorry for any inconvience or any time consumed to help me with this problem. Thank you very much.

~Eric