# Can't find the velocity

1. Mar 5, 2009

### physicsnewb7

1. The problem statement, all variables and given/known data
This problem arose in my dynamics class. It says that a ball is thrown from a 1.5 m height at an angle of 3 degrees upward. it reaches a wall 6 meters away from where the ball is thrown, and hits the wall at a height of .97 m. The question asks what is the velocity?

2. Relevant equations
y=v(t)+1/2a(t^2)
v(y)=v+at

3. The attempt at a solution
I tried to get the x and y velocity components but just cant seem to do it. I can't seem to understand how to get a velocity when there is no given amount of time. please help

2. Mar 5, 2009

### thepopasmurf

haven't tried working it out but you can make the question simpler by shifting the vertical height values down by 1.5m. That way, the ball starts at the origin and the ending height is (.97 - 1.5) = -.53m

This gets rid of some messy stuff. About your time question, I haven't tried it yet but because you are working on two dimension of motion, you can probably equate the equations and t is not required.

PS.
velocity vectors:
ucos3 in the x direction
usin3 in the y direction

3. Mar 5, 2009

### physicsnewb7

how am I supposed to calculate the x,y velocity components ucos(3) and usin(3) when i cant find the normal velocity.

4. Mar 5, 2009

### thepopasmurf

you leave them as usin3 and ucos3 (or 0.05234u and 0.99863u)

the method is this basically:
sx = 6 when sy = -0.53 (I got -0.53 from my previous post)
so this means that:

6 = ucos(3)t and -0.53 = usin(3)t -.5gt^2

t = 6/ucos(3) and you sub this into the other equation, then there is a lot of donkey work to solve for u

5. Mar 5, 2009

### physicsnewb7

thanks appreciate the help!