MATL, 35 34 characters

Z"42Gi:qi!+lGqE!/tYo-KM!*|QYoZ}X]( 

Try it online!

Explanation

This takes care of the reflections using the formulas

nx = 2*(Nx-1) px = abs(t/nx-round(t/nx))*nx 

where t is a time variable that increases in steps of 1, Nx is the maximum size along the x direction and px is the computed x position. Analogous formulas are used for the y coordinate.

A char matrix of spaces is initially created, and then it is filled with character * at the computed x and y positions.

Z" % Implicitly take [width height]. Push char matrix of spaces with that size 42 % Push 42. ASCII for character `*` G % Push first input again, that is, [width height] i % Take number of movements, M :q % Range [0 1 ... M-1] i! % Take initial position [x y]. Transpose into column vector + % Add element-wise with broadcast. Gives 2-row vector lG % Push first input, [width height] qE! % Transform into column vector [2*width-2; 2*height-2] / % Divide element-wise with broadcast tYo % Duplicate. Round to closest integer - % Subtract KM! % Push [2*width-2; 2*height-2] again * % Multiply element-wise with broadcast | % Absolute value Q % Add 1 Yo % Round to closest integer Z} % Separate 2-row array into two row arrays. These contain row and column % indices, respectively, of the ball positions X] % Convert row and colummn indices to linear index ( % Assign 42 to those positions of the matrix of spaces 

Python 180 172 171 163 145 Bytes

(-2 Bytes with Python3)

def r(m,n,x,y,t): s=([' ']*m+['\n'])*n;a=b=1 for c in'o'*(t-1)+'x':s[x+y*m+y],x,y=c,x+a,y+b;a*=2*(0<x<m-1)-1;b*=2*(0<y<n-1)-1 print''.join(s) 

To save 2 more bytes with Python3 (thanks to shooqie) replace second line with

 s=[*' '*m,'\n']*n 

Had to use a list of characters because python does not support string item assignement. The x+y*m+y part is from x+y*(m+1) because of the '\n' add each line end

Edit1: right after posting, saw how to get rid of the overwriting 'x'

Edit2: space after print

Edit3: using < and > instead ==, also tabs for 2nd indentation

Edit4: Replaced the if statements with assignements, daisy-chaining less-than and placing the loop in one line (thanks to RootTwo)

JavaScript, 293 279

Nothing too clever going on here.

EDIT: Saved 14 characters by initialising the 2D array differently.

p=prompt().split(' ') a=[];for(i=0;i++<p[1];)a.push(new Array(~~p[0]).fill(' ')) x=~~p[2];y=~~p[3] d=e=1 for(i=0;i++<p[4];){a[y][x]=i==p[4]?'x':'o' if(x<1&&d<0||x>p[0]-2&&d>0)d*=-1 if(y<1&&e<0||y>p[1]-2&&e>0)e*=-1 x+=d;y+=e}o='' for(i=0;i++<p[1];)o+=a[i-1].join('')+'\n' alert(o)

R, 156 bytes/chars

As an unnamed function. It's quite long but I quite like the method. There's probably a better way to do the 1 to w to 1 vector.

function(w,h,x,y,i)cat(c(' ','o')[1:(w*h)%in%(rep(c(c(1:w),c((w-1):2)),i)[x:((i=i-1)+x)+1]+(rep(c(c(1:h),c((h-1):2)),i)*w-w)[y:(i+y)+1])*1+1],fill=w,sep='') 

Brief explanation of function internals

cat( # output to STDOUT c(' ','o') # vector of characters [1:(w*h)%in%( # number of characters in the board rep(c(c(1:w),c((w-1):2)),i)[x:((i=i-1)+x)+1]+ # oscillating vector of x to w to 1 to x (rep(c(c(1:h),c((h-1):2)),i)*w-w)[y:(i+y)+1] # similar to above but in steps of w *1+1] # cause a vector of 1 or 2 to pick char ,fill=w,sep='') # print width of w, no separator 

Like the other current answers this doesn't output the border, but I will add one if required.

Javascript ES6 202 181 177 chars

(w,h,x,y,s)=>(a=(r=Array)(h).fill().map(_=>r(w).fill` `),eval("for(vx=vy=1;s--;x+=vx,y+=vy,a){a[y][x]=0;if(!x||x==w-1)vx*=-1;if(!y||y==h-1)vy*=-1;}"),a.map(r=>r.join``).join` `) 

Not very golfed, I take a another look at it once I get home.

C++11, 355 Bytes

Takes input string as command line arguments

#include <iostream> #define A(n,i) int n=atoi(a[i]); int main(int,char**a) { A(w,1)A(h,2)A(x,3)A(y,4)A(n,5) int d=1,r=1; auto b=new int64_t[h]; while(--n>1){ b[y]|=1<<x; if(r)x++;else x--; if(d)y++;else y--; if(x==0||x==w-1)r=!r; if(y==0||y==h-1)d=!d; } for(int i=0;i<h;++i){ for(int j=0;j<w;++j){ std::cout<<(x==j&&y==i?'x':b[i]&(1<<j)?'o':' '); } std::cout<<'\n'; } } 

Don't worry about the newlines, they are already removed from the count.

Storing the rows as 64-Bit integer is okay since maximum width is 50.

Q, 94 Bytes (kx.com)

{[w;h;i;j;n]./[(1+h,w)#" ";+{z#y_+\0,-1 1@{x=2*_x%2}'&(-_-(z+y)%x)#x}.'(h,j,n;w,i,n);:;"*"]}.

Defines a lambda with parameters w (width), h (height), i (initial x), j (initial y), and n (number of movements).

Lambda is delimited with {}

Final . (after }) allows to use a sequence as argument. Standard function invocation is f[arg;arg;..;arg], but we can also invoque as f . sequence. An integer sequence is v v .. v, so we can invoque as f . w h i j n. If we define f:{..}. we can invoque as f w h i j n (as required by the question)

Solution uses 'K mode' of Q. K is the functional inner core of Q (Q is more readable, verbose and with a lot of additional libraries, but K is compact). To execute in K mode we must use scripts with k extension (vs q extension for q mode). Interactive interpreter starts in Q mode (-> q) prompt). We can load code from k script, and execute test directly on interactive interpreter (even in q mode, as shown in test). There is also a k mode at interactive interpreter level, but it's not worth in this case.

TEST

NOTE.- to avoid display cluttering, we assume f name associated to lambda (f:{..}.)

q)f 35 15 0 0 49 "* * " " * * * " " * * * " " * * * " " * * * " " * * *" " * * * " " * * * " " * * * " " * * * " " * * * " " * * * " " * * * * " " * * * * " " * * * * " " * * " q)f 35 15 4 2 49 " * " " * * " " * * * " " * * *" " * * * " " * * * " " * * * " " * * * " " * * * " " * * * " " * * * * " " * * * * " " * * * " " * * * * " " * * * * " " * * " 

EXPLANATION

We define a lambda for 1D movement that accepts args [max;init;n] (legal positions 0..max-1, initial pos init, n movements) and returns a sequence of positions (after computing rebounds). Later apply that lambda to y and x axes (with args [h;j;n] and [w;i;n] respectively).

• Initially assume starting at 0 pos. We calculate number of rebounds to accomplish n movements, and generate a sequence with repeated increasing numbers (0 0 ...0; 1 1 .. 1; 2 2 .. 2; ..).

• Each subsequence has w items, so there are rebounds just between subsequences. For each number determine if is even, and sustitutes even results by 1 and even results by -1. (1 1 .. 1; -1 -1 .. -1; 1 1 .. 1; ..)

• We join subsequences and calculate accumulated sums starting with 0 -> 0 1 2 3 .. 35 34 33 32 .. 0 1 2 ..

• finally we drop starting i movements and take exactly n movements

Last, we define a (h x w) char matrix filled with blanks, and for each position (resulting of applying 1D lambda on y and x) we replace character with "*"