# Projectile Velocity

1. Jun 4, 2007

### scw287

A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.00 m/s and her body makes an angle of 81.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
magnitude m/s
direction °

Here's what I've tried to do: I tried to solve for the initial velocity by using the kinematics equation v^2=vo^2+2*a*x
vo=1.174<<initial velocity

Now I need to find the magnitude and the direction, I think you use tan-1 and pythagorean theorem but I don't really understand the component method very well. Could you please help me get on the right track for this problem. All help will greatly be appreciated.

2. Jun 4, 2007

### ice109

do you know the position function?

3. Jun 4, 2007

### scw287

no I haven't heard of the position function.

4. Jun 4, 2007

### ice109

do you know the kinematic equations?

5. Jun 4, 2007

### andrevdh

How did you get the initial y velocity 1.174 m/s?

6. Jun 4, 2007

### scw287

Yes I know the kinematics equations x=vot+a/2at^2
x=1/2(vo+v)t
v=vo+at
v^2=vo2+2ax

I got 1.74m/s for the initial velocity by using the fourth equation. I am not positive that I did it right. Not sure what to do next!

7. Jun 4, 2007

### skeeter

in the y-direction ...

v_{fy}^2 = v_{oy}^2 - 2g(Delta y)

v_{oy} = sqrt[v_{fy}^2 + 2g(Delta y)]

v_{oy} = sqrt[(9*sin(81))^2 + 2(9.8)(-3)] = approx 4.5 m/s

velocity in the x-direction is a constant ...

v_x = 9*cos(81) m/s

you now have both x and y components of the initial velocity ... finish up.

8. Jun 4, 2007

### scw287

Thanks for all of your help!!

9. Jun 5, 2007

### andrevdh

Yes, I thought that you did use the fourth equation, but I got a different answer. I was hoping that you would show more exactly how you susbtituted to get to it so that we can compare notes.