Solution: waist AB=AC= 15/cos30? = 10√3, then when PC=AC/cos30? =( 10√3)/(√3/2)=20, PA⊥AC, and so on.

When PC= 10, PA⊥AB,

So when point P moves for 5 seconds 10 second, PA is perpendicular to the waist of △ABC.

Another simple solution is:

Because PA⊥AC, then ∠BAP=30? , so PBA is an isosceles triangle, that is, PB=PA, at this time, PA=PC÷2.

Therefore, when PB= 10, PA⊥AC,

Similarly, when PB=20, PA⊥AB

therefore ...