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A body is dropped from a height 122.5 m. If its stopped after 3 seconds and again released the further time of descent is (g=9.8 m/s²)

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4 s

3 s

5 s

2 s

answer is C.

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Detailed Solution

During 3s time height of fall $h = \frac{1}{2} \times 9.8 \times 3^{2}$

= 4. 9 × 9 = 44.1

Remaining height = 122.5 - 44.1 = 68.4

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 68.4}{9.8}} = 4 s$

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