Question

A body is dropped from a height of 122.5m. If it is stopped after 3 seconds and again released. The further time of descent is ($g=9.8m{{s}^{-2}}$)
A. 2s
B. 3s
C. 4s
D. 5s

Answer 1

Hint: The motion of any object is supported and explained very clearly with the help of newton’s equations of motion. There are three equations of motion. These equations give a relation between the initial and final velocity, distance covered by the body, acceleration, and time.
As per the given data,
The initial height at which the object is placed is 122.5m above the ground
The object is stopped In between after a time interval of 3s.

Formula used: $s=u+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step answer:
The journey of the object of reaching the ground can be divided into two sections according to the stops in the path,
Case 1: (when the object starts falling from 122.5m)
The length covered by the body before reaching the point at which the body is stopped after the time duration is 3s can be calculated with the help of the 2nd equation of motion.
By putting the values the length covered to reach the point is,
$\begin{align}
  & s=u+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow s=0+\dfrac{1}{2}9.8\times {{(3)}^{2}} \\
 & \Rightarrow s=\dfrac{88.1}{2} \\
 & \Rightarrow s=44.1m \\
\end{align}$
The height at which the object is stopped is,
$\begin{align}
  & 122.5-44.1 \\
 & =78.4m \\
\end{align}$
Case 2: (When the object is stopped at a height of 78.4m and starts to fall again)
The time taken by the body to reach the ground will be,
$\begin{align}
  & s=u+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow 78.4=0+\dfrac{9.8}{2}{{(t)}^{2}} \\
 & \Rightarrow {{t}^{2}}=\dfrac{78.4}{4.9} \\
 & \Rightarrow {{t}^{2}}=16 \\
 & \Rightarrow t=4s \\
\end{align}$

So, the correct answer is “Option C”.

Note: When the body is stopped after a time duration of 3s it loses its speed. When it is again released it starts its journey again with a zero initial velocity. When a body is in free-falling only acceleration due to gravity works on the body.