Ok, if some one were to hit a ball at a heigth of 3 ft and it was about 25 ft high at 380 ft away from home plate how far would it go? The only other given I have is that the ball had a velocity of 115 to 120 mph off the bat.

Wait, I don't understand all that you're saying.

The ball is hit at 3 ft from the ground, is that right?
The maximum height the ball reaches is 25 ft?
The ball is hit 380 ft away from the home plate?

If that is so, then:
The ball lost all its vertical speed at 25 ft high, or (25-3) ft from the projected height.

From: v^2 = u^2 + 2as

0^2 = u^2 + 2(-32.2)(22)

You get the initial vertical speed as being 37.6 ft/s

If you make use of the formula for range, that is R = \frac{v^2 sin2\theta}{g}

Then, you will need to find the angle of projection.

37.6 ft/s is the vertical component. The angle of projection will therefore be obtained from:

115 sin \theta = 37.6

\theta = 19.1^o

Plug that in your formula to find the horizontal distance covered 3ft above the ground. This should be enough to know whether it's a home run or not.

To find the exact total distance covered, find the time for the ball to hit the ground, using:

s = ut + \frac12 gt^2

-3 = (37.6)t + \frac12 (-32.2)t^2

Solve for t, then use:

s = vt\ cos\theta

To find the total horizontal distance.

EDIT: Note that I used 115 ft/s, and not 115 mph

I would apprach this differently. I am assuming that we know the ball barely clears the fence 380 feet from home plate. The question is how much further does the ball go before hitting the ground?

Of course this depends on the arc the ball takes. If it's a high shot that barely clears the 25 foot fence, the total distance is not much more than 380 ft. But if it's a line drive hit it may carry way beyond 380 ft.

Start with the known facts:

y = y_i + Vy t - 1/2 gt^2
x = Vx t

if Vx = 120 MPH = 176 ft/s, then

380 ft = 176 ft/s x t
t = 2.16 seconds

Put this into the equation for y for when the ball clears the fence:

y = 25 ft = 3 + v_y (2.16 s) - 1/2 (32.2) 2.16 s^2
v_y = 44.75 ft/s

Now determine when does y = 0:

y = 0 = 3 + 44.75 ft/s x t - 1/2 (32.2) t^2

Solve for t using the quadratic equation: t = 2.86s

So the total distance carried is:

x = 176 ft /s x 2.86s = 504 ft

Thanks for your responses. I hate to admit that I don't understand some of the symbols used in the formulas so I ask you to break them down a little further.
I will add that this home run was a screaming line drive that according to witnesses cleared the 15 ft fence by about 10 ft (hence 25 ft) near the 376 ft mark. And that it still kept rising beyond the fence. Exaggerated reports said that the ball went 613 ft. It is humanly impossible to hit a line drive ball that far at that angle. The velocity of the ball off the bat would have to be 200 mph! Big league sluggers are known to hit the long ball
Around 110 - 115 mph and we assume they are hit waist high or 3 ft above the ground. Of course there are exceptions possibly beyond 120 mph like Babe Ruth or Mickey Mantle but 200 mph is absurd. Please elaborate further for me. Thanks

Raymond - the key to this calculation is your estimate of the ball being hit at between 115 MPH and 120 MPH. That pretty much dictates how far the ball carries. The claim that the ball was still rising at 380 feet is specious. To do that the speed of the ball off the bat would have had to be at least 225 MPH! And that's assuming no air resistance - add the effect of air resistance and the initial speed of the bat must be greater. I think the claim that the ball was still rising at 380 feet out is a clear exageration, but one made for effect to reinforce that it was indeed a mighty hit.