Question

A body is falling freely from a point $A$ at a certain height from the ground and passes through point $B,C$ and $D$ (vertically as shown) so that $BC=CD$. The time taken by the particle to move from $B$ to $C$ is $2$ seconds and from $C$ to $D$ is $1$ second. Time taken to move from $A$ to $B$ in second is

• A. $0.6$
• B. $0.5$
• C. $0.2$
• D. $0.4$

Answer 1

Answer: (B)

$0.5$

$BC=2V+\frac{g}{2}\times 2^2$

$BC=2V+2g$
$2BC=3V+\frac{1}{2}g\times 3^2$

$2BC=3V+\frac{9g}{2}$
$4V+4g=3V+\frac{9g}{2}$

$V=\frac{9g}{2}-4g$
$=g/2$
$A\to B$

$V=u+at$
$\frac{g}{2}=0+g\times t$

$t=1/2$