Answer
a) $h=16.5m$
b) $F=2.95N$
Work Step by Step
a) If there is no air resistance, the energies are conserved: $$E_f-E_0=0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=0$$ $$\Big(\frac{1}{2}v_f^2+gh_f\Big)-\Big(\frac{1}{2}v_0^2+gh_0\Big)=0$$
The projectile is shot straight up with $v_0=18m/s$, and it reaches its highest point when $v_f=0$. The starting point is $h_0=0$. So, $$gh_f-\frac{1}{2}v_0^2=0$$ $$h_f=\frac{v_0^2}{2g}=16.5m$$
b) When there is air resistance, the energies are not conversed, so according to the work-energy theorem: $$W_{nc}=E_f-E_0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=W_{nc}$$
Again, since $v_f=0$ and $h_0=0$, we have $$W_{nc}=mgh_f-\frac{1}{2}mv_0^2$$
We have $h_f=11.8m$, $v_0=18m/s$ and $m=0.75kg$, so $W_{nc}=-34.77J$
Air resistance force opposes the projectile's motion along its way so $\cos\theta=\cos180=-1$ and the distance $s=h_f=11.8m$
$$W_{nc}=-Fs$$ $$F=\frac{W_{nc}}{-s}=2.95N$$
