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Grade 9Mechanics

In the last second of its freefall a body cover ¾ of its total path . Then the height from which the body is released will be ?

Varun Gandhi

10 Years agoGrade 9

1 Answer

ADITI SINGH

10 Years ago

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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