Projectile Question involving xyz coordinate plane

[icanletyougo]

Homework Statement

Please help me graph this on xyz coordinate plane
A golfer strikes the ball of the tee at x=0;y=0;z=0. The hole center is located at a distance of 150m north(+x) and 75m west(+y) of the tee. The upper edge of the hole is 10m below(-z) the height of the tee.
If the the ball is launched from the tee at a 30 degree angle respect to the horizontal, what is the minimum launch velocity need to make the ball land into the hole.
Thank

Homework Equations

The Attempt at a Solution

I have no idea how to graph it.

 

[icanletyougo]

Homework Statement

Please help me graph this on xyz coordinate plane
A golfer strikes the ball of the tee at x=0;y=0;z=0. The hole center is located at a distance of 150m north(+x) and 75m west(+y) of the tee. The upper edge of the hole is 10m below(-z) the height of the tee.
If the the ball is launched from the tee at a 30 degree angle respect to the horizontal, what is the minimum launch velocity need to make the ball land into the hole.
Thank

Homework Equations

The Attempt at a Solution

I have no idea how to graph it.

I think I need to find the distance from the tee to the hole. And then from that, use the x-component to figure out t, and then substitute t into the y-component eq. to find initial velocity. But I'm not sure...Can anyone help me?
Thank you very much

 

[AdrianMay]

I suggest you just forget the horizontal plane and think about it in 2D. You can reconstruct the X,Y axes when you're finished.

Please refer to my post in this thread: https://www.physicsforums.com/showthread.php?t=477477 about how to construct the parabola.

In your case, it's a bit different from that thread because you're given the vertical height and you have to solve for vertical and horizontal velocity, whereas the guy in that thread had Vx and Vy and had to solve for the height. That's not a big difference though. You both got the initial angle for free and that's the important bit. Your graph looks scary because the ball is going to sink below where you hit it, but as long as you watch the signs that won't be a problem.

As for what kind of picture they want you to draw, I have no idea.

Adrian.

 

[icanletyougo]

I suggest you just forget the horizontal plane and think about it in 2D. You can reconstruct the X,Y axes when you're finished.

Please refer to my post in this thread: https://www.physicsforums.com/showthread.php?t=477477 about how to construct the parabola.

In your case, it's a bit different from that thread because you're given the vertical height and you have to solve for vertical and horizontal velocity, whereas the guy in that thread had Vx and Vy and had to solve for the height. That's not a big difference though. You both got the initial angle for free and that's the important bit. Your graph looks scary because the ball is going to sink below where you hit it, but as long as you watch the signs that won't be a problem.

As for what kind of picture they want you to draw, I have no idea.

Adrian.

Thanks Adrian.
This is my picture of the 2D motion
[PLAIN]http://img190.imageshack.us/img190/6691/24050103.jpg
So I have a question,
Do I need the OA length in order to find the time the ball travel?
This is my work...but I'm not sure it is correct...
OA = 167.7 m
From x-component we have
167.7 = v*cos(30)*t
t = 167.7/(v*cos(30))
From the y-component we have
-10 = (v*sin(30))*(t) -.5at^2
-10 = (v*sin(30))*(167.7/(v*cos(30)) - .5*(9.81)*(167.7/(v*cos(30))^2
From that I solve for v= 41.49 m/s
IS that right ?
Thank you so much