Why does it jump a line of code in the try block?

It jumps over it because an exception was thrown, and it's not just one line. Any code that is placed after an exception throwing line in a try block will be skipped.

The purpose of a try/catch block is that the try block will essentially be on the lookout for exceptions thrown by the code within it. And once one occurs will automatically pass the exception to whatever catch block has been setup to deal with the exception. Once that happens the job of the try/catch block is done so the remaining code is not ran.

If you wanted the isAcceptable line to run regardless of whether the applyGuess threw an error or not then you could simply place it outside of the try block like this:

try { 
    isHit = game.applyGuess(guess);
} catch(IllegalArgumentException iae) { 
    System.out.println("%s. Please try again. %n", iae.getMessage()); 
}
isAcceptable = true;

Generally you should only place code that might throw exceptions and code that should only run if no exception occurred from that code within the try block. Any code you want to run regardless should be placed outside try blocks.

In the example you provided the isAcceptable variable is likely there to signal if the applyGuess actually ran without issues. So having it inside of the try block makes sense.

After an exception was thrown any code after that line will be skipped. So only if isHit = game.applyGuess(guess) is valid isAcceptable will be true.

class TryCatch {
    public static void main(String args[]) {
        try {
            System.out.println("You can see me.");
            int number = 50 / 0;
            System.out.println("You can't see me.");
        } catch (ArithmeticException ae) {
            System.out.println("Error: You can't divide a number by zero");
        }
    }
}

Output :

You can see me.

Error: You can't divide a number by zero