A body is falling freely from a point $A$ at a certain height from the ground and passes through points $B, C$ and $D$ (vertically as shown below) so that $B C=C D$. The time taken by the particle to move from $B$ to $C$ is $2 \mathrm{~s}$ and from $C$ to $D$ is 1 s. Time taken to move from $A$ to $B$ in seconds is

• A

  0.6
• Correct answer
  B

  0.5
• C

  0.2
• D

  0.4

Solution

Let velocity of the particle at point $B$ be $v$. Now, $\quad B C=2 v+\frac{1}{2} g \times(2)^2$ or, $$ B C=2 v+2 g $$ Similarly, $$ 2 B C=3 v+\frac{1}{2} \times g \times(3)^2 $$ $$ \Rightarrow \quad 2 B C=3 v+\frac{9 g}{2} $$ From Eq. (i), $$ \begin{array}{rlrl} 2(2 v+2 g) & =3 v+\frac{9 g}{2} \\ \Rightarrow & 4 v+4 g & =3 v+\frac{9 g}{2} \\ \Rightarrow & 4 v-3 v & =\frac{9 g}{2}-4 g \\ \Rightarrow & & v & =\frac{g}{2} \end{array} $$ From point $A$ to $B$, $$ v=u+a t $$ $$ \begin{aligned} \Rightarrow & & \frac{g}{2}=0+g \times t \\ \Rightarrow & & t=\frac{1}{2} \text { second } \\ \Rightarrow & & t=0.5 \mathrm{~s} \end{aligned} $$

Hence, the correct answer is option: 'B'.