According to the Hint, it is enough to check the sign of v = h ′ {\displaystyle v=h'} at t = π 6 {\displaystyle t={\frac {\pi }{6}}} .
From the part (b), we have v ( π 6 ) = 0.1 > 0 {\displaystyle v\left({\frac {\pi }{6}}\right)=0.1>0} , and therefore, h {\displaystyle h} is increasing at t = π 6 {\displaystyle t={\frac {\pi }{6}}} .
Answer : m o v i n g u p {\displaystyle \color {blue}moving\ up}
