Answer
$d$ = 4 m
Work Step by Step
Using the period equation of SHM we get the mass 2 by
$$m_{2}=\dfrac{k T^{2}}{4 \pi^{2}}= \dfrac{(1208.5) (0.14)^{2}}{4 \pi^{2}}= 0.6 \mathrm{kg}$$
From the conservation law we get the final velocity of mass 1 by
\begin{gather*}
m_{1}\left(v_{1 i}-v_{1 f}\right)=m_{2}\left(v_{1 i}+v_{1 f}\right)\\
\left|v_{1 f}\right|=\left|\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right| v_{1 i}\\
\left|v_{1 f}\right|=\left|\frac{0.2 \mathrm{kg}-0.6 \mathrm{kg}}{0.2 \mathrm{kg}+0.6 \mathrm{kg}}\right|(8 \mathrm{m} / \mathrm{s})\\
\left|v_{1 f}\right| = 4 \,\text{m/s}
\end{gather*}
Hence, we use this value of the speed as its the initial speed of the projectile $v_o = v_1f$ to get the distance $d$
$$d=v_o t= v_{o} \sqrt{\frac{2 h}{g}}=(4 \mathrm{m} / \mathrm{s}) \sqrt{\frac{2(4.9 \mathrm{m})}{9.8 \mathrm{m} / \mathrm{s}^{2}}}=4 \mathrm{~m}$$
