Question

A particle at rest falls under gravity such that it travels 53.9m in the last second of its journey. Total time of fall is

Answer 1

Explanation:

just used the formula of 'displacement in n th second

Answer 2

Answer:

The total time of fall is 7 sec.

Explanation:

Given that,

Distance d = 53.9 m

Using equation of motion

$S_{n}=u+\dfrac{1}{2}g{2n-1}$

Where, S = distance

u = initial velocity

n = time

Put the value in the equation

$53.9=0+\dfrac{1}{2}\times9.8\times(2(t-1)-1)$

$53.9=4.9(2t-3)$

$53.9+14.7=9.8t$

$t =\dfrac{53.9+14.7}{9.8}$

$t =7\ sec$

Hence, The total time of fall is 7 sec.