A) 1s
B) 3s
C) 4 s
D) None of these
Correct Answer: A
Solution :
Acceleration \[{{a}_{\max }}=\frac{T-mg}{m}\] \[=\frac{(20-10)g}{10}=g\] During half time, body is accelerate from rest and during next half time decelerates with same magnitude to come to rest. \[\frac{h}{2}=\frac{1}{2}{{a}_{\max }}t_{1}^{2}\] \[{{t}_{1}}=\sqrt{\frac{h}{{{a}_{\max }}}}=1s\]
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