# Really Need Help here with 2D Projectile Motion

1. Oct 17, 2011

### Tronic

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

A diver springs upward from a board that is 3m above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75 degrees with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

Initial velocity = V1 , Final Velocity = V2

Assuming Down as positive and up as negative
Y direction X direction
Dy = 3m. Dx = Unknown.
Ay = 9.8 Ax = 0
v2y = Sin 75 8.90 v2x = Cos 75 8.90
v1 = unknown v1x = v2x ( since there is no accelaration ! )

2. Relevant equations

3. The attempt at a solution

Since V1x = v2x , We do not need to calculate V1x since its already 2.3 ( cos 75 8.90 )

V2Y = ( Sin 75 8.90 ) = 8.60

(8.60) = V1Y^2 + (2x9.8x3)
(8.60)^2-(2x9.8x3) = V1y^2
3.89 = V1y^2

Since 3.89 is positive, it should be pointing down. But when i make a vector with y point down and x pointing to the right, after using pythagoras theorem and tan , i get the wrong degree. However, if i assume down to be as negative and up to be as positive, then my direction is correct.
I have always been told by my teachers that its upto me what i take as negative and positive. I want to know why im not being able to get the answer when i take down as positive and up as negative. either way i get my Y value as positive whether i take positive down or up , so how do i know which one is right.

Help will really be appreciated. Thanks.

2. Oct 17, 2011