Question

If a freely balling body covers half of its total distance in the last second of the its journey. Find its time of fall...

Sir, please let me know how to solve this step by step...

Answer 1

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