A block of mass \(m = 10\,kg\) rests on a horizontal table. The coefficient of friction between the block and the table is \(0.05.\) When hit by a bullet of mass \(50\,g\) moving with speed \(v,\) that gets embedded in it, the block moves and comes to stop after moving a distance of \(2\,m\) on the table. If a freely falling object were to acquire speed \(\frac {v}{10}\) after being dropped from height \(H,\) then neglecting energy losses and taking \(g = 10\,ms^{-2},\) the value of \(H\) is close to ................. \(\mathrm{km}\)

\(\begin{gathered}
  a = \frac{f}{{M + m}} = \frac{{\mu \left( {M + m} \right)g}}{{(M + m)}} = \mu g \hfill \\
   = 0.05 \times 10 = 0.5\,m{s^{ - 2}} \hfill \\
  {V_0} = \frac{{Initial\,momentum\,}}{{\left( {M + m} \right)}} = \frac{{0.05V}}{{10.05}} \hfill \\
  m = 50g\,\,\,\,\,\,\,\,\,\,\,\,M = 10\,kg \hfill \\ 
\end{gathered} \)

\({\left( {\frac{{0.05v}}{{10.05}}} \right)^2} = 2 \times 0.5 \times 2\)

Solving we get \(v = 201\sqrt 2 \) object falling from height \(H.\)

\(\begin{gathered}
  \frac{V}{{10}} = \sqrt {2gh}  \hfill \\
  \frac{{201\sqrt 2 }}{{10}} = \sqrt {2 \times 10 \times H}  \hfill \\ 
\end{gathered} \)

\(H = 40\; m = 0.04\; km\)

\(\begin{gathered}
  {v^2} - {u^2} = 2as \hfill \\
  0 - {u^2} = 2as \hfill \\
  {u^2} = 2as \hfill \\ 
\end{gathered} \)