A) \[6\sec \]
B) \[5\sec \]
C) \[4\sec \]
D) \[3\sec \]
Correct Answer: B
Solution :
The distance covered in\[3\sec \], \[{{s}_{3}}=ut+\frac{1}{2}g{{t}^{2}}\] or \[{{s}_{3}}=0\times t+\frac{1}{2}9.8\times {{3}^{2}}\] or \[{{s}_{3}}=44.1m\] The distance covered in last second \[4.9(2t-1)=44.1\] \[2t-1=\frac{44.1}{4.9}=9\] or \[2t=9+1=10\] or \[t=5\sec \] As both the distances are equal then \[4.9(2t-1)=44.1\] \[2t-1=\frac{44.1}{4.9}=9\] or \[2t=9+1=10\] or \[t=5\sec \]
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