Question

if a freely falling body covers half of its distance in the last second of the journey . find its time of fall

Answer 1

Answer:

ANSWER IS 2 root 2

Explanation:

we know that Sn=u+a/2(2n-1) and u = 0

do Sn=5(2t-1)

S=5t^2

S/2=5t^2

S/2t^2=5(2t-1)

t^2=4t-2

t^2=4t+2=0

t=4+ root under16-4 (1×2)/2

4+2under root 2/2

so answer =2root under 2

Answer 2

Answer:

The time of fall for the given problem will be 2 + √2 seconds.

Explanation:

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 = ($\frac{1}{2}$) gt²

= ($\frac{1}{2}$) gn² this will be actual distance travelled in n seconds.

H(n-1) = ($\frac{1}{2}$) g (n - 1)² this will be distance travelled in n - 1 seconds.

So, the distance travelled in the nth second (last second) h = Hn - H(n - 1)

= ($\frac{1}{2}$)g {n² - (n - 1)²}

= ($\frac{1}{2}$) g (2n - 1)

As per the question h = ($\frac{1}{2}$) Hn

⇒ ($\frac{1}{2}$) g (2n - 1) = ($\frac{1}{4}$) gn²

=> 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, and will be rejected.

So the time of journey (Time of fall) n = 2 + √2 seconds.

Thus, the free-fall time is the time it takes for a body to collapse by its own gravitational influence if no other forces are present to prevent it. It is crucial in determining the timescale for a broad range of astrophysical events.