JavaScript (ES6), 174 bytes

Takes input in curling currying syntax ([x, y])(a), where x and y are the 0-indexed coordinates of the starting position and a[ ] is a matrix of integers, with 0 = ice, 1 = wall, 2 = sand, 3 = hole and 4 = water

Returns 0 if there's no solution.

p=>a=>(r=F=([x,y],n,R=a[y],c=R[x])=>R[c&(R[x]=4)|n>=r||[-1,0,1,2].map(d=>(g=_=>(k=a[v=Y,Y+=d%2][h=X,X+=~-d%2])||g())(X=x,Y=y)>3?0:k>2?r=-~n:F(k>1?[X,Y]:[h,v],-~n)),x]=c)(p)|r 

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Commented

p => a => ( // given the starting position p[] and the matrix a[] r = // r = best result, initialized to a non-numeric value F = ( // F = recursive function taking: [x, y], // (x, y) = current position n, // n = number of shots, initially undefined R = a[y], // R = current row in the matrix c = R[x] // c = value of the current cell ) => // R[ // this will update R[x] once the inner code is executed c & (R[x] = 4) | // set the current cell to 4 (water); abort if it was n >= r || // already set to 4 or n is greater than or equal to r [-1, 0, 1, 2].map(d => // otherwise, for each direction d: (g = _ => ( // g = recursive function performing the shot by k = a[ // saving a backup (h, v) of (X, Y) v = Y, Y += d % 2][ // and updating (X, Y) until we reach a cell h = X, X += ~-d % 2]) // whose value k is not 0 (ice) || g() // )(X = x, Y = y) // initial call to g() with (X, Y) = (x, y) > 3 ? // if k = 4 (water -> fail): 0 // abort immediately : // else: k > 2 ? // if k = 3 (hole -> success): r = -~n // set r to n + 1 : // else: F( // do a recursive call to F(): k > 1 ? // if k = 2 (sand): [X, Y] // start the next shots from the last cell : // else (wall): [h, v], // start from the last ice cell -~n // increment the number of shots ) // end of recursive call ), x // end of map(); x = actual index used to access R[] ] = c // restore the value of the current cell to c )(p) | r // initial call to F() at the starting position; return r 

Python 3, 273 bytes

def p(g,c,d,k=0): 	while 1>k:c+=d;k=g.get(c,9) 	return-(k==2)or c-d*(k==3) def f(g): 	c={q for q in g if g.get(q,9)>4};I=0;s=[c] 	while all(g.get(q,9)-4for q in c): 		c={k for k in{p(g,k,1j**q)for k in c for q in range(4)}if-~k} 		if c in s:return-1 		s+=[c];I+=1 	return I 

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-41 bytes thanks to ovs
-1 byte thanks to Jonathan Frech

C#, 461 418 bytes

This is just a non-competitive reference implementation to (hopefully) revive this challenge:

Golfed by Kevin Cruijssen

int P(string[]C){int w=C[0].Length,i=0,l=c.Length;var c=string.Join("",C);var h=new int[l];for(var n=new List<int>();i<l;n.Add(i++))h[i]=c[i]!='@'?int.MaxValue:0;for(i=1;;i++){var t=n;n=new List<int>();foreach(int x in t){foreach(int d in new[]{-1,1,-w,w}){for(int j=x+d;c[j]==' ';j+=d);if(c[j]=='#'&h[j-d]>s){h[j-d]=s;n.Add(j-d);}if(c[j]=='.'&h[j]>s){h[j]=s;n.Add(j);}if(c[j]=='o')return s;}}if(n.Count<1)return -1;}} 

Ungolfed

int IceGolf(string[] course) { // Width of the course int w = course[0].Length; // Course as single string var c = string.Join("", course); // Array of hits per field var hits = new int[c.Length]; // Fields to continue from var nextRound = new List<int>(); // Initialize hits for (int i = 0; i < hits.Length; i++) { if (c[i] != '@') // All fields start with a high value hits[i] = Int32.MaxValue; else { // Puck field starts with 0 hits[i] = 0; nextRound.Add(i); } } for (int s = 1; ; s++) { // clear the fields that will be used in the next iteration var thisRound = nextRound; nextRound = new List<int>(); foreach (int i in thisRound) { // test all 4 directions foreach (int d in new[] { -1, 1, -w, w }) { int j = i+d; // ICE - slide along while (c[j] == ' ') j += d; // WALL - stop on previous field if (c[j] == '#' && hits[j-d] > s) { hits[j-d] = s; nextRound.Add(j-d); } // SAND - stop if (c[j] == '.' && hits[j] > s) { hits[j] = s; nextRound.Add(j); } // HOLE return strikes if (c[j] == 'o') return s; } } // No possible path found if (nextRound.Count == 0) return -1; } } 

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