$A$ ball is projected vertically upward with an initial velocity of $50 \; ms^{-1}$ at $t = 0 \; s$. At $t = 2 \; s$,another ball is projected vertically upward with the same velocity. At $t = \dots \; s$,the second ball will meet the first ball $(g = 10 \; ms^{-2})$.
- A$6$
- B$5$
- C$4$
- D$3$
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