Time taken by the particle to reach a height $h$ from the surface of Earth is obtained from $h=u t-\frac{1}{2} g t^2$
Here, $u=9 \mathrm{~m} / \mathrm{s}, h=4 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned} & \therefore \quad 4=9 t-\frac{1}{2} \times 10 \times t^2 \\ & 4=9 t-5 t^2 \quad \text { or } 5 t^2-9 t+4=0 \\ & 5 t^2-5 t-4 t+4=0 \\ & \Rightarrow \quad 5 t(t-1)-4(t-1)=0 \\ & \therefore \quad t=1 \mathrm{~s} \text { or } t=\frac{4}{5} \mathrm{~s}\end{aligned}$
Hence, the time taken by the particle to reach a height of $4 \mathrm{~m}$ from the surface second time is $1 \mathrm{~s}$.