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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

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


 

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5 Answers

Aman Bansal
592 Points
11 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
7 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
7 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
4 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 .
Kushagra Madhukar
askIITians Faculty 628 Points
4 years ago
Dear student,
Please find the answer to your question.
 
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

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