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A particle at rest, falls under gravity $(g = 9.8 m/s^2)$ such that it travels 53.9 m in the last second of its journey . Total time of fall is
(A) 4 s
(B) 5 s
(C) 6 s
(D) 7 s

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Last updated date: 17th Apr 2024
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Answer
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Hint: The particle at a height \[h\] at rest (initial velocity is zero), moves under gravity and reaches the ground at time t seconds. The distance travelled in the last second is given (position). Since the position and time of the moving body is concerned (with zero initial velocity), the second law of equation should be applied.

Formula used:
Using the second equation of motion, under gravity g,
\[v = ut + \dfrac{1}{2}g{t^2}\]
The initial and final velocity is denoted by ‘u’ and ‘v’ respectively. And, ‘t’ is the time taken to cover the full distance.

Complete step by step answer:
Initially a particle is at rest, say, at a height h from the ground.
Since the particle is at rest, the distance covered is 0
$velocity = \dfrac{{displacement}}{{time}}$
And, the initial velocity $u = 0m{s^{ - 1}}$
Distance travelled in last second $d = 53.9m$
Given, gravitational force \[g = 9.8m{s^{ - 2}}\]
Let the total time taken by the particle to fall (from the height h to the ground) = $t$
Since the particle started from rest,
$u = 0$
At time $t = t - 1$ seconds,
Therefore, distance travelled by the moving particle in \[\left( {t - 1} \right)\] seconds = $S$
\[\Rightarrow S = 0 + \dfrac{1}{2}g{\left( {t - 1} \right)^2}\]
\[\Rightarrow S = \dfrac{1}{2}g{\left( {t - 1} \right)^2}\]
Similarly , distance travelled in t seconds \[ = 0 + \dfrac{1}{2}g{t^2}\]
\[\Rightarrow S = \dfrac{1}{2}g{t^2}\]
Height h = distance travelled in (t-1) seconds + distance travelled in last second
\[\begin{array}{*{20}{l}}
  {\Rightarrow \dfrac{1}{2}g{t^2} = \dfrac{1}{2}g{{\left( {t - 1} \right)}^2} + 53.9} \\
  { \Rightarrow \dfrac{1}{2}g{{ }}\left( {{t^2}-{t^2} + 2t - 1} \right){{ }} = 53.9} \\
  { \Rightarrow g\left( {2t-1} \right){{ }} = 2 \times 53.9} \\
  { \Rightarrow 9.8\left( {2t-1} \right){{ }} = {{ 107}}{{.8}}} \\
  { \Rightarrow 2t-1 = \dfrac{{53.9}}{{4.9}} = {{ }}11} \\
  {\therefore t = 6\sec }
\end{array}\]
Total time of fall = 6 seconds

Hence option (C) is the correct one.

Note: When the Velocity and time of a moving body are related (and for non-zero initial velocity), First equation of motion should be applied. It goes like this
\[{{v}} = u + at\]
Acceleration due to gravity (\[g = 9.8{{ }}m{s^{ - 2}}\]) is denoted by ‘a’. And, ‘u’ and ‘v’ denote the initial and final velocity respectively.