Suppose that both the balls meet at a distance $h$ from ball-l after time $t$ from the start as shown in the figure,
For downward motion of ball-1, from second equation of the motion,
$$
\begin{aligned}
& h=u t+\frac{1}{2} g t^2 \\
& h=0 \times t+\frac{1}{2} \times 10 t^2 \quad \ldots(i) \\
& h=5 t^2 \quad[\because u=0]
\end{aligned}
$$




For upward motion of ball-2, from second equation of the motion,
$$
h=v t-\frac{1}{2} g t^2
$$
Given, speed of ball-2,v $=14 \mathrm{~m} / \mathrm{s}$, acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
Substituting these values in Eq. (iii), we get
$$
\begin{aligned}
& 21-h=14 t-\frac{1}{2} \times 10 t^2 \\
\Rightarrow \quad & 21-h=14 t-5 t^2
\end{aligned}
$$
Adding Eq. (ii) and (iv), we get
$$
\begin{aligned}
& 21=14 t \\
& \Rightarrow \quad t=\frac{3}{2} \mathrm{~s}=1.5 \mathrm{~s} \\
&
\end{aligned}
$$
$\therefore$ From Eq. (ii), we get
$$
\begin{aligned}
h & =5 \times(1.5)^2=11.25 \mathrm{~m} \\
& =\frac{45}{4} \mathrm{~m}
\end{aligned}
$$
Therefore, the ball-l will have dropped $\frac{45}{4} \mathrm{~m}$ when it passes ball-2.