Question

A particle at rest falls under gravity (g = 9.8m/s) such that it travels 53.9m in last second of the journey total time of the fall is:

Answer 1

Hint: In order to solve this question we have to use one of the equation of the motion for the distance while applying three different conditions, one is the distance in total time t, second one is the distance covered in (t-1) second and the third one time taken for the distance of 53.9 m by using this we can derive total time of the fall for the particle.

Formula used:
\[\begin{align}
& s=ut+\dfrac{1}{2}a{{t}^{2}} \\
& {{s}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right) \\
\end{align}\]

Complete step by step answer:
Let’s take the total time taken by the particle to fall is t.
Now we know that equation of the motion for the distance s and the time t
\[s=ut+\dfrac{1}{2}a{{t}^{2}}....\left( 1 \right)\]
Here particle full form rest hence initial velocity u = 0 and it is falling under the gravity therefore the acceleration a will be gravitational acceleration ‘g’ in equation (1)
S = total distance travelled by the particle
u = initial velocity of the particle
t = total time
a = acceleration of the particle
Now substitute values in the equation (1)
$\begin{align}
  & \Rightarrow {{s}_{t}}=0+\dfrac{1}{2}g{{t}^{2}} \\
 & \therefore {{s}_{t}}=\dfrac{1}{2}g{{t}^{2}}....\left( 2 \right) \\
\end{align}$
Now distance travelled by the particle in (t-1) s.
$\begin{align}
  & {{s}_{t-1}}=0+\dfrac{1}{2}g{{\left( t-1 \right)}^{2}} \\
 & \therefore {{s}_{t-1}}=\dfrac{1}{2}g{{\left( t-1 \right)}^{2}}...\left( 3 \right) \\
\end{align}$
Now the distance travelled by the particle in the last one second.
${{s}_{t-1}}=53.9m....\left( 4 \right)$
Now from the equation (2) (3) and (4) we can write that the total time taken by the particle is,
${{s}_{t}}={{s}_{t-1}}+{{s}_{t=1}}...\left( 5 \right)$
Now let’s substitute all the values in the equation (5)
$\begin{align}
  & \dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g{{\left( t-1 \right)}^{2}}+53.9 \\
 & \Rightarrow \dfrac{1}{2}g{{t}^{2}}-\dfrac{1}{2}g\left( {{t}^{2}}-2t+1 \right)=53.9 \\
 & \Rightarrow \dfrac{1}{2}\times 9.8\left( {{t}^{2}}-\left( {{t}^{2}}-2t+1 \right) \right)=53.9 \\
 & \Rightarrow 4.9\left( {{t}^{2}}-\left( {{t}^{2}}-2t+1 \right) \right)=53.9 \\
 & \Rightarrow 2t-1=\dfrac{53.9}{4.9} \\
 & \Rightarrow 2t-1=11 \\
 & \Rightarrow 2t=12 \\
 & \therefore t=6s \\
\end{align}$

Therefore option (c) 6s is correct.

Note:
There is another short cut method to solve this question we know that formula for the distance at n second is,
\[{{s}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)\]
Replace n with t, and a with ‘g
Here,
$\begin{align}
  & {{s}_{n}}=53.9m \\
 & u=0m/s \\
 & a=9.8m/{{s}^{2}}\left( g \right) \\
 & n=t \\
\end{align}$
Substitute all the values in equation
\[\begin{align}
  & \Rightarrow 53.9=0+\dfrac{1}{2}\times 9.8\left( 2t-1 \right) \\
 & \Rightarrow 11=2t-1 \\
 & \Rightarrow 2t=12 \\
 & \therefore t=6s \\
\end{align}\]