Given that, the height achieved in $1 \mathrm{~s}, h=25 \mathrm{~m}$ Let $u$ be the initial velocity in upward direction. By equation of motion,
$$
h=u t-\frac{1}{2} g t^2
$$


$$
\begin{aligned}
& 25=u(1)-\frac{1}{2}(10)(1)^2 \\
& 25=u-5 \Rightarrow u=30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Final velocity of ball at maximum height become instantly zero.
$\therefore$ Time taken to reach at maximum height is calculated as
$$
v=u-g t \Rightarrow 0=u-10 t \Rightarrow t=\frac{30}{10}=3 \mathrm{~s}
$$
Now, distance travelled in $2 \mathrm{~s}=$ Displacement in $2 \mathrm{~s}$ By equation of motion,
$$
d_1=s_1=u t-\frac{1}{2} g t^2
$$
$$
\begin{aligned}
& =30(2)-\frac{1}{2} \times 10 \times(2)^2 \\
& =60-5 \times 4=60-20=40 \mathrm{~m}
\end{aligned}
$$
Maximum height achieved by ball in $(3 \mathrm{~s})$
$$
\begin{aligned}
H & =u t-\frac{1}{2} g t^2=30(3)-\frac{1}{2} \times 10(3)^2 \\
& =90-\frac{10}{2} \times 9 \\
& =90-45=45 \mathrm{~m}
\end{aligned}
$$
Displacement after $4 \mathrm{~s}$,
$$
\begin{aligned}
S_2 & =u t-\frac{1}{2} g t^2=30 \times 4-\frac{1}{2} \times 10(4)^2 \\
& =120-(5 \times 16) \\
& =120-80=40 \mathrm{~m}
\end{aligned}
$$
Now, distance travelled, $d_2=H+\left(H-S_2\right)$
$$
\begin{aligned}
& =45+(45-40) \\
& =45+5=50 \mathrm{~m}
\end{aligned}
$$
Ratio, $\frac{d_1}{d_2}=\frac{40}{50}=\frac{4}{5}$