\begin{align*} I & =\int _{0}^{t}Fdt\\ & =\int _{0}^{t}mgdt\\ & =mgt\\ & =\left ( 1\right ) \left ( 9.81\right ) \left ( 41\right ) \\ & =402.21\text{ N-s} \end{align*}
\[ -e=\frac{V_{B}^{+}-V_{A}^{+}}{V_{B}^{-}-V_{A}^{-}}\] Where \(B\) is the wall. Hence \(V_{B}^{+}=V_{B}^{-}=0\) since wall do not move. Therefore\begin{align*} -e & =\frac{-V_{A}^{+}}{-V_{A}^{-}}\\ & =\frac{-\left ( -1.10\right ) }{-\left ( +4.48\right ) }\\ & =\frac{1.10}{-4.48}\\ & =-0.24554 \end{align*}
Hence \[ e=0.245\,54 \]
\[ mv_{A}^{-}+\int _{0}^{0.29}F_{av}dt=mv_{A}^{+}\] But \(m=1\) then\begin{align*} \int _{0}^{0.29}F_{av}dt & =v_{A}^{+}-v_{A}^{-}\\ & =-1.47-5.13\\ & =-6.6 \end{align*}
Hence \begin{align*} F_{av}\left ( 0.29\right ) & =-6.6\\ F_{av} & =-\frac{6.6}{0.29}\\ & =-22.759 \end{align*}
The magnitude is \(22.759\) N. The negative sign, since force is in negative \(x\) direction.
Energy lost is
\begin{align*} \Delta & =\frac{1}{2}m\left ( v_{A}^{-}\right ) ^{2}-\frac{1}{2}m\left ( v_{A}^{+}\right ) ^{2}\\ & =\frac{1}{2}\left ( 6.4\right ) ^{2}-\frac{1}{2}\left ( 0.12\right ) ^{2}\\ & =20.473\text{ J} \end{align*}
Let \(A\) be the fallcon and \(B\) be the drone. Hence
\begin{align*} m_{A}v_{A}^{-}+m_{B}v_{B}^{-} & =\left ( m_{A}+m_{B}\right ) v^{+}\\ \left ( 1.023\right ) \left ( 107\right ) +0 & =\left ( 1.023+1\right ) v^{+}\\ v^{+} & =\frac{\left ( 1.023\right ) \left ( 107\right ) }{2.023}\\ & =54.108\text{ m/s} \end{align*}