a body covers half of its total height in the last second of its free fall calculate i)duration of fall ii)height from which body was dropped

Dear Student,

Time of journey (free fall) = t = n seconds
Height of free fall = H meters ; 
initial velocity = 0 ; acceleration due to gravity = g = 9.8 m/s²
Hn = (1/2) g t² = (1/2) g n² = distance travelled in n seconds
H(n-1) = (1/2) g (n - 1)² = distance travelled in n - 1 seconds
Distance 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 seconds
Height of fall = (1/2) g (2 + √2)² = (1/2)(9.8)(4 + 2 + 4√2) = 9.8*(3 + 2√2) meters
(Calculate by taking √2 = 1.414

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