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 (Neglect air friction and take $g = 10 m / s^{2}$ )

3 s

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5 s

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7 s

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9 s

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Solution

The correct option is B $5 s$
Distance travelled in the $n^{t h}$ second $= u + \frac{g}{2} (2 n - 1)$

Distance travelled in the first 3 seconds:
$= u t + \frac{1}{2} g t^{2}$
$= 0 \times 3 + \frac{1}{2} \times 10 \times 3^{2} = 45 m$

Distance travelled in last second = $S_{n} = 45 = 5 (2 n - 1) \Rightarrow n = 5$

Hence, the body has fallen for a time of $5 s$ .

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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 (Neglect air friction and take g=10 m/s2)

The correct option is B 5 sDistance travelled in the nth second =u+g2(2n−1) Distance travelled in the first 3 seconds: =ut+12gt2 =0×3+12×10×32=45 m Distance travelled in last second = Sn=45=5(2n−1)⇒n=5 Hence, the body has fallen for a time of 5 s.

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 (Neglect air friction and take $g = 10 m / s^{2}$ )