# 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

Aman Bansal
592 Points
10 years ago

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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Afzaal
23 Points
6 years ago
A body is failling from rest at a certain hight.The distance covered by body in first 3 second is equal to the distance coverd by body in last second before striking the ground.How long body remains in air
Kajal singh
11 Points
5 years ago
T2=2h/g: (3)Square=2h/10: h=45mS=u+1/2a(2n-1) ((foumula to find displacement in one second))-45=0+1/2*-10(2t-1) where n= tseconds(2t-1) =45*2/10T=5seconds
mahesh kumar
15 Points
3 years ago
u=0
half distance in last second
[h/2= g/2[2n-1] ….....1
fall freely from rest
h= 1/2gn^2 ….........2
from equation 1and 2 we find
g/2[2n-1]=g/4n^2
2n-1=n^2/2
4n-2=  n^2
n^2-4n+2=0
[n-2]^2-2 =0
[n-2]^2 =2
n-2 =2^1/2
n=2^1/2+2 .
2 years ago
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) = 57.12 m

Thanks and regards,
Kushagra