A) \[2s,3s\]
B) \[1.5s,\text{ }2.5s\]
C) \[1.2s,\text{ }3.2s\]
D) \[3.2s,1.3s\]
Correct Answer: D
Solution :
Take the y-axis to be upward, at\[{{v}_{0}}=0\] and solve\[y={{v}_{0}}t-\frac{1}{2}g{{t}^{2}}\]for t \[\therefore \] \[t=\sqrt{-2y/g}\] Here for \[y=-50\text{ }m\] So, \[t=\sqrt{-\frac{2(50)}{9.8}}=3.2s\] Here, \[y=-100m\] \[t=\sqrt{-\frac{2(-100)}{9.8}}=4.5\,s\] Hence, the difference is the time taken to fall the second 50 m. \[=4.5-3.2=1.3s\]
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