Two bodies are thrown vertically upward, with the same initial velocity of 98 m/s but 4 sec apart. How long after the first one is thrown when they meet ?
Text Solution
Verified by ExpertsThe correct answer is:
C
o (t + 4)
S 1 = S 2
⇒ ut – 
gt 2 = u (t + 4) – 
g (t + 4) 2
⇒ ut – 
gt 2 = ut + 4u – 
g [t2 + 16 + 8t]
⇒ 4u – 
[16 + 8t] = 0
⇒ t = 8
Time of meeting = t + 4 = 12 sec
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