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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
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Verified Answer
The correct answer is:
0.5
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 $B$,
$$
v=u+a t
$$
$$
\begin{aligned}
\Rightarrow & & \frac{g}{2}=0+g \times t \\
\Rightarrow & & t=\frac{1}{2} \text { second } \\
\Rightarrow & & t=0.5 \mathrm{~s}
\end{aligned}
$$
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