Question

A particle at rest , falls under gravity ( g=9.8m/s) such that it travels 53.9 in the last second of it's journey. Total time of fall is?​

Answer 1

A particle is in rest at height h from the ground, now particle starts to fall under gravity such that it travels 53.9m in the last second.

Let total time taken by particle to fall = T sec

then, last sec , particle travels 53.9 m

now, height travels by particle in (T - 1) seconds , s = ut + 1/2at²

here, u = 0, a = g

or, s = 0 + 1/2(g) × (T - 1)²

or, s = 1/2 g(T - 1)² ....(1)

and similarly distance covered by particle in T seconds

h = 1/2 gT².....(2)

we know, h = distance covered in (T - 1) sec + distance covered in last second

or, 1/2 gT² = 1/2g(T - 1)² + 53.9

or, 1/2g(T² - T² + 2T - 1) = 53.9

or, 1/2g(2T - 1) = 53.9

or, 1/2 × 9.8(2T - 1) = 53.9

or, 2T - 1 = 53.9/4.9 = 11

or, T = 6sec

hence, total time of fall = 6 sec.

shortcut :

$S_n=u+\frac{1}{2}a(2n-1)$

where, Sn = 53.9m, u = 0, a = g = 9.8m/s² and n = T

then, T = 6sec

Answer 2

Answer:

Total time of fall is 6 sec.

Explanation:

Given that,

Initial velocity u= 0

Acceleration due to gravity g = 9.8 m/s

Distance d= 53.9 m

We need to calculate the time,

Using equation of motion

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

Where, u = initial velocity

n = time

g = acceleration due to gravity

s = distance

Put the all value in equation (I)

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

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

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

$2t=12$

$t=6\ sec$

Hence, Total time of fall is 6 sec.