Phsyics Kinematics: cannonball shot straight up

[okgo]

http://www.screencast.com/users/trinhn812/folders/Jing/media/5c3b51ea-aed8-4321-964c-126b49a3c1dc

F/m = 0.3m/s^2
distance in x direction= 1/2(a)(T^2)

By placing v(y) =0 and v(initial)=100 I found the time to be 20 sec.
But my answer is 60m. So 1020-60=960m

Wouldn't the ball misss the target by more than 960 meters? Why are the options ranging from 15-60. Perhaps I'm missing the point here. The acutal answer is 31.2

 

[jfy4]

http://www.screencast.com/users/trinhn812/folders/Jing/media/5c3b51ea-aed8-4321-964c-126b49a3c1dc

F/m = 0.3m/s^2
distance in x direction= 1/2(a)(T^2)

By placing v(y) =0 and v(initial)=100 I found the time to be 20 sec.
But my answer is 60m. So 1020-60=960m

Wouldn't the ball misss the target by more than 960 meters? Why are the options ranging from 15-60. Perhaps I'm missing the point here. The acutal answer is 31.2

here is my math for it.

[tex]v_{y}=v\sin\theta[/tex]
[tex]v_{x}=v\cos\theta[/tex]

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

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

 

[okgo]

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

 

[okgo]

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

 

[jfy4]

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

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

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.

 

[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