Question

A particle falling from rest under gravity covers a
height H in 5 seconds. If it continues falling then
next distance will be covered in approximately​

Answer 1

Answer:

$\huge\bigstar\underline\mathfrak{Solution}$

We have given:

• u (initial velocity) =0m/s
• t (Time)= 5seconds
• Height=h

By applying formula:

$s = ut + \frac{1}{2} g {t}^{2}$

Since height is h and initial velocity is 0m/s, we get:

$= > h = \frac{1}{2} g {t}^{2}$

$= > h = \frac{1}{2} \times 10 \times {5}^{2}$

$= > h = 125m$

Starting from a position labeled as 0, and starting with a velocity of 0, the distance D traveled in time T is proportional to the square of the time. (s=(1/2)at² may be familiar to you.)

So if the time to fall distance H is 5 sec, then the total time to fall 2H is √2×(5sec) and the time to fall 3H is √3×(5sec), and so on. Those are total times, from the start, so to get the time spent between H and 2H, subtract the first 5 sec to find (√2-1)×(5sec) .

That is your will be 6.07 second approximately after solved (√2-1)×5 second