Question

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) so that $BC=CD$. The time taken by the particle to move from $B$ to $C$ is $2$ seconds and from $C$ to $D$ $1$ second. Time taken to move from $A$ to $B$ in seconds is

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

Answer 1

The correct option is B $0.5$

Velocity of the body at point A $u_A=0$
Let the velocity of body at points B and C be $u_B$ and $u_C$ respectively.
Let $BC=CD=S$
For journey $A$ to $B$ :
Using $u_B=u_A+g{t}_{AB}$
where $t_{AB}$ is the time taken to move from A to B.
$\therefore$ $u_B=10t_{AB}$ .......(1) $(\because u_A=0)$
For journey $B$ to $C$ :
Using $u_C=u_B+g{t}_{BC}$
$\therefore$ $u_C=u_B+20$
Using $S=u_B{t}_{BC}+\frac{10}{2}{t}_{BC}^2$
where $t_{BC}=2$
We get $S=2u_B+5(2{)}^2=2u_A+20$ .....(2)
For journey $C$ to $D$ :
Using $S=u_C{t}_{CD}+\frac{10}{2}{t}_{CD}^2$
where $t_{CD}=1$
We get $S=u_C+5(1{)}^2=u_C+5$ .....(3)
Equation (2) - (1), we get $0=2u_B-u_C+15$
$⟹2u_B=u_C-15$
Or $2u_B=u_B+20-15$
$⟹u_B=5$
From (1), we get $5=10t_{AB}$
$⟹t_{AB}=0.5s$