A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of ( in seconds upto two decimals)

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Solution

Distance covered in first three seconds = Distance covered in last second
$\Rightarrow \frac{1}{2} g (3)^{2} = \frac{g}{2} (2 t - 1) \Rightarrow t = 5 s$

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A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of ( in seconds upto two decimals)

Distance covered in first three seconds = Distance covered in last second ⇒12g(3)2=g2(2t−1)⇒t=5 s

Distance covered in first three seconds = Distance covered in last second
$\Rightarrow \frac{1}{2} g (3)^{2} = \frac{g}{2} (2 t - 1) \Rightarrow t = 5 s$