A) 6 sec
B) 8 sec
C) 10 sec
D) 12 sec
Correct Answer: D
Solution :
Let t be the time of flight of the first body after meeting, then \[(t-4)\]sec will be the time of flight of the second body. Since \[{{h}_{1}}={{h}_{2}}\] \[\therefore \]\[98t-\frac{1}{2}g{{t}^{2}}=98(t-4)-\frac{1}{2}g{{(t-4)}^{2}}\] On solving, we get \[t=12\]seconds
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