c
\(YX\)

\(v = 400 m/s\)

\(Mass \,before\, explosion = m\)

\(490\, m\, and\, velocity \,v = 200\, m/s\, (vertically)\)

Momentum before explosion

 \(=\)Momentum after explosion

\(\begin{array}{l}
m \times 200\hat j\, = \frac{m}{2} \times 400\hat j + \frac{m}{2}\,v\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{m}{2}\left( {400\,\hat j + v} \right)
\end{array}\)

\(\begin{array}{l}
 \Rightarrow \,400\,\hat j - 400\hat j = v\\
\therefore \,\,v = 0
\end{array}\)

i.e., the velocity of the other part of the mass, \(v = 0\)

Let time taken to reach the earth by this part be \(t\)

Applying formula, \(h = ut +\) \(\frac{1}{2}g{t^2}\)

\(\begin{array}{l}
490 = 0 + \frac{1}{2} \times 9.8 \times {t^2}\\
 \Rightarrow \,{t^2} = \frac{{980}}{{9.8}} = 100\\
\therefore \,\,t = \sqrt {100}  = 10\sec 
\end{array}\)