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`        If a freely falling body covers half of its total distance in the last second of its journey.find it'd time of fall`
one year ago

```							Dear student Time of journey (free fall) = t = n secondsHeight of free fall = H meters ; initial velocity = 0 ; acceleration due to gravity = g = 9.8 m/s²Hn = (1/2) g t² = (1/2) g n² = distance travelled in n secondsH(n-1) = (1/2) g (n - 1)² = distance travelled in n - 1 secondsDistance travelled in the n th (last) second = h = Hn - H(n-1)= (1/2)g {n² - (n - 1)²} = (1/2) g (2n - 1)As per the question : h = (1/2) Hn(1/2) g (2n - 1) = (1/4) g n²=> 2n - 1 = n²/2=> n² = 4n - 2=> n² - 4n + 2 = 0=> (n - 2)² - 2 = 0 => (n - 2)² = 2=> n - 2 = ± √2=> n = 2 ± √2=> n = (2 + √2) or (2 - √2)n = 2 - √2, being less than 1, is rejected.Time of journey = n = 2 + √2 secondsHeight of fall = (1/2) g (2 + √2)² = (1/2)(9.8)(4 + 2 + 4√2) = 9.8*(3 + 2√2) meters
```
one year ago
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