Phsyics Kinematics: cannonball shot straight up

1. Feb 28, 2010

okgo

2. Feb 28, 2010

jfy4

here is my math for it.

$$v_{y}=v\sin\theta$$
$$v_{x}=v\cos\theta$$

$$t=\frac{v\sin\theta}{g}$$
to get the total time in the air multiply by two, becuase that time is only half way.
$$x=v\cos\left(\theta\right) t$$
plug in t
$$x=\frac{2v^2\sin\theta \cos\theta}{g}$$

that should get you the distance traveled... i think but look over my math.

3. Mar 1, 2010

okgo

Oh. I'm having trouble with the angle though. Not sure what it would be.

4. Mar 1, 2010

okgo

and isn't there acceleartion in the x direction too?

5. Mar 1, 2010

jfy4

From what i saw in the problem there were not any more forces than just gravity... and gravity only works in the y direction. your angle is the one given in the problem, 45.

6. Mar 1, 2010

okgo

Oh I was talking about question 23. Hehe I already solved for question 22. Oh well. My exam is in 20 min. Dumdumdum. So it's okay. Thanks for the help though