4)
the arrow touches the ground when the heigth is 0
so we need to replace h=0 and solve for t
[tex]\begin{gathered} h=17t-5t^2 \\ 0=17t-5t^2 \\ 0=t(17-5t) \\ 0=17-5t \\ 5t=17 \\ t=\frac{17}{5}=3.4 \end{gathered}[/tex]the arrow tuches the ground at 3.4 seconds
5)
[tex]F=m\times a[/tex]replacing
[tex]185=m\times13.7[/tex]solve for m
[tex]\begin{gathered} m=\frac{185}{13.7} \\ \\ m=13.50 \end{gathered}[/tex]the mass is 13.50Kg