Balls A and B are released from rest from roof of a building at t = 0 and t = 2 s, respectively. The ball A strikes with ground and comes back with same speed. After some time, both balls A and B meet with each other at a height of 55 m from the ground. If the height of building is 60n metre, then the value of n is ___________.

$\begin{array}{l}\text{ If they meet at }t=t_0\text{. }\\ \therefore h-55=\frac{1}{2}g{\left(t_0-2\right)}^2\text{ (for ball B) .........(i)}\end{array}$

where, h is height of the building. The velocity of each ball at the meeting point is v₀.

$\therefore v_0=g\left(t_0-2\right)$

Let the time taken by ball A to go from meeting point ground is t₁

$\begin{array}{l}\therefore t_0=t_0-2+2t_1\\ \text{ Or }{t}_1=1\mathrm{s}\end{array}$

$\begin{array}{l}\therefore 55=v_0{t}_1+\frac{1}{2}g{t}_1^2\\ \text{ Or }55=g\left(t_0-2\right)t_1+\frac{1}{2}g{t}_1^2\end{array}$

$\begin{array}{l}\text{ Or }5.5=\left(t_0-2\right)t_1+\frac{t_1^2}{2}=t_0-2+\frac{1}{2}\\ \text{ Or }5=t_0-2\end{array}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{t}_0=7\mathrm{s}\\ \because \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h-55=\frac{1}{2}g{\left(t_0-2\right)}^2\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h-55=5(7-2{)}^2=125\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h=55+125=180\mathrm{m}=60 n\mathrm{m}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n=3\end{array}$