Answer

Verified

35.7k+ views

**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.

Recently Updated Pages

To get a maximum current in an external resistance class 1 physics JEE_Main

f a body travels with constant acceleration which of class 1 physics JEE_Main

A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main

If the beams of electrons and protons move parallel class 1 physics JEE_Main

Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Other Pages

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Chloroform reacts with oxygen in the presence of light class 12 chemistry JEE_Main

Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main

The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main