Moving all the terms to the left, we have the equation $ x^2-14x+y^2-48y=0 $. Completing the square on the quadratic in $ x $, we add $ (14/2)^2=49 $ to both sides. Completing the square on the quadratic in $ y $, we add $ (48/2)^2=576 $ to both sides. We have the equation \[(x^2-14x+49)+(y^2-48y+576)=625 \Rightarrow (x-7)^2+(y-24)^2=625\] Rearranging, we have $ (y-24)^2=625-(x-7)^2 $. Taking the square root and solving for $ y $, we get $ y=\pm \sqrt{625-(x-7)^2}+24 $. Since $ \sqrt{625-(x-7)^2} $ is always nonnegative, the maximum value of $ y $ is achieved when we use a positive sign in front of the square root. Now, we want the largest possible value of the square root. In other words, we want to maximize $ 625-(x-7)^2 $. Since $ (x-7)^2 $ is always nonnegative, $ 625-(x-7)^2 $ is maximized when $ (x-7)^2=0 $ or when $ x=7 $. At this point, $ 625-(x-7)^2=625 $ and $ y=\sqrt{625}+24=49 $. Thus, the maximum $ y $ value is $ \boxed{49} $. --OR-- Similar to the solution above, we can complete the square to get the equation $ (x-7)^2+(y-24)^2=625 $. This equation describes a circle with center at $ (7,24) $ and radius $ \sqrt{625}=25 $. The maximum value of $ y $ is achieved at the point on the top of the circle, which is located at $ (7,24+25)=(7,49) $. Thus, the maximum value of $ y $ is $ \boxed{49} $.