If a freely falling body covers half of it's total distance in the last second of it's journey. Find it's time of fall. Pls write the steps and don't just give the answer
Answers
Answer:hi
Explanation: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