### Arrow

It is best to first answer part (b). The average velocity of the arrow while decelerating is $v_{y,av}=\frac{1}{2}\left(0+260\frac{m}{s}\right)=130\frac{m}{s}.$
The time for the arrow to travel 9 cm (0.09 m) is $t=\frac{0.09m}{130\frac{m}{s}}=6.92\times10^{-4}s$
(a) The acceleration of the arrow is then $a_{y}=(v_{y}-v_{0y})/t=\left(0-\left(260\frac{m}{s}\right)/6.92\times10^{-4}s\right)=-27.5\times10^{4}\frac{m}{s^{2}}$