Question

7. A particle at rest, falls under gravity
such that it travels 53.9 m in las
journey. Total time of fall is
(1) 45
(2) 5s
(3) 65
(4) 7s​

Answer 1

S=53.9m
g=9.8 g/cm^3
u=0
S=ut+(1/2)at^2
53.9=0+(1/2)9.8*t^2
53.9=0+4.9t^2
t^2=53.9/4.9
t^2=11
t=root over 11

Answer 2

Answer:

Total time of fall is 6 s.

3 option is correct.

Explanation:

Let us assume that the particle takes time t to reach the ground.

The distance in last second $s_{t}= 53.9\ m$

Using equation of motion

$s_{n}=ut+\dfrac{1}{2}g(2n-1)$

Where, n = t

u = initial velocity

g = acceleration due to gravity

Put the value in the equation

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

$2t-1=\dfrac{53.9}{4.9}$

$2t =11+1=12$

$t=6\ s$

Hence, Total time of fall is 6 s.