using grep for particular lines in text files

A quick one liner would be:

$ awk '$5 == '0' || $5 == '15' ||  $5 == '30' ||  $5 == '45' {print}' test.txt

As fedorqui mentioned in the comment below and even quicker one liner is:

$ awk '$5 ~ /^(0|15|30|45)$/' test.txt

Syntax explaned:

$5  - the desired column from the text.
~   - the match symbol... matching was between the two slashes (/).
^/$ - the regex symbols indicating the beginning and end of the field.
|   - the or specifier for either item in the "()" group.

The header is filtered out by only accepting lines that match the items following the match ~ criteria.

It is perfectly do-able in grep, although awk is the go-to tool for field separated data.

With grep:

grep -E '^10[[:blank:]]+([^[:blank:]]+[[:blank:]]+){3}(0|15|30|45)[[:blank:]]' file.txt

• Matching 10 at the start, ([^[:blank:]]+[[:blank:]]+){3} matches the next 3 fields and then matching 5th field for the desired ones

If you use PCRE (-P), you can replace [:blank:] with \s, looks good on eyes:

grep -P '^10\s+([^\s]+\s+){3}(0|15|30|45)\s' file.txt     

Example:

$ cat file.txt                                        
    2              METEOROLOGICAL DATA                         VERSION /      
    8    PR    TD    HR    ZW    ZT    WD    WS    RI      # / TYPES OF   MOD/TYPE/ACC 
 3979316.8350  1050313.7180  4857065.7030      592.1910 PR SENSOR POS XYZ/H           
                                                           END OF HEADER 
10  1  1  0  0 15  927.9    4.3   99.1                
10  1  1  0  1 15  927.9    4.3   99.1                
10  1  1  0  2 15  927.9    4.3   99.1                
10  1  1  0 15 15  927.9    4.2   99.1               
10  1  1  0 16 15  927.9    4.2   99.0                
10  1  1  0 30 15  927.7    4.1   99.1                
10  1  1  0 31 15  927.7    4.1   99.1                
10  1  1  0 45 15  927.5    4.1   99.1                
10  1  1  0 46 15  927.5    4.0   99.1                
10  1  1  1  0 15  927.4    4.1   99.1                
10  1  1  1  1 15  927.4    4.1   99.1    

$ grep -E '^10[[:blank:]]+([^[:blank:]]+[[:blank:]]+){3}(0|15|30|45)[[:blank:]]' file.txt
10  1  1  0  0 15  927.9    4.3   99.1                
10  1  1  0 15 15  927.9    4.2   99.1               
10  1  1  0 30 15  927.7    4.1   99.1                
10  1  1  0 45 15  927.5    4.1   99.1                
10  1  1  1  0 15  927.4    4.1   99.1                

$ grep -P '^10\s+([^\s]+\s+){3}(0|15|30|45)\s' file.txt                         
10  1  1  0  0 15  927.9    4.3   99.1                
10  1  1  0 15 15  927.9    4.2   99.1               
10  1  1  0 30 15  927.7    4.1   99.1                
10  1  1  0 45 15  927.5    4.1   99.1                
10  1  1  1  0 15  927.4    4.1   99.1        

You can do this using sed and awk:
(assuming your subject text is in a file called example.txt)

sed -n '/END OF HEADER/,${
/END OF HEADER/d
p
}' example.txt | awk '($5 == 0 || $5 == 15 || $5 == 30 || $5 == 45)'

_{Attribution - the awk command was taken from steeldriver's comment}

Explanation

Sed

• ignore everything up to /END OF HEADER/
• /END OF HEADER/d means delete the line containing /END OF HEADER/ out of the pattern space, and because its no longer in the pattern space, the next command p won't do anything
• p - print each line as it gets pulled into the pattern space

Awk

• if column 5 is 0 or 15 or 30 or 45 then print it