Get a Ruling • Re: Bal in water hazard

quincy wrote:

OempaS wrote:Player plays his bal from the tee into a water hazard. He drops a bal and plays it.
He realises that he dropped the bal by mistake inside the hazard.
He drops another bal now in the right place  outside the hazard and plays it onto the green. He completes the hole with another two puts.
What is the ruling? And how many strokes must he write on his scorecard?

The ball played from within the water hazard is in play with an additional two-stroke-penalty for playing from a wrong place and the player should have continued play with it provided the breach was not a serious one.  When he abandoned it and put another ball into play  he proceeded under stroke and distance incurring the penalty stroke under R. 27-1 and an additional two-stroke-penalty for a breach of that Rule, again for playing from a wrong place.

According to my math he has made  five strokes and must add six penalty strokes for a score of 11.

quincy

I agree with Quincy, subject to one qualification.  If, after having played the substituted ball mistakenly from the hazard, the player could not see his ball (or no-one informed him precisely where it is), then I am in full agreement with Quincy's explanation.  As that ball is now in play (ie it has now become the 'original' ball in terms of what he has to do next - proceed under R13-1) and he cannot see it or know precisely where it is and he proceeds under an inapplicable rule (he thinks he is correcting and proceeding under R26), the only rule that can be allocated by the Committee to this circumstance is R27-1 (1SP) plus the additional 2SP for wrong place (as described by Quincy). 
On the other hand, if he could see his ball or was informed precisely where it is, then when he proceeds under the inapplicable rule the Committee must allocate R13-1 to judge his subsequent actions.  In such a situation, he has incurred a further 2 SP rather than 3, for a total of 5SP and 5 strokes, a total score of 10.  The authority for the latter 2SP is D18-2/8 - see the last paragraph of the answer.

Statistics: Posted by Thom — Today, 05:07

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