Question

A,B,C and D are points in a vertical line such that AB=BC=CD. If a body falls from rest at A, then the times of descent through AB, BC and CD are in the ratio
A. $1:\sqrt{2}:\sqrt{3}$
B. $\sqrt{2}:\sqrt{3}:1$
C. $\sqrt{3}:1:\sqrt{2}$
D. $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2}-1)$

Answer 1

Hint: We know that the distances are equal to each other and hence from a reference point A, they will be in the ratio of 1:2:3. Using the second equation of kinematics for constant acceleration for these distances, we will find the time taken by the body to fall from point A to B, B to C and C to D and find the ratio.

Complete step by step answer:
We know the second equation of kinematics for constant acceleration as there is linear motion occurring, the equation is;
\[h = ut + \dfrac{1}{2}a{t^2}\]
Here, h is the distance travelled, u is the initial velocity, t is the time taken and a is the constant acceleration. In our case, initial velocity is zero and the acceleration is due to the acceleration of gravity. Thus, the time taken is given by:
\[t = \sqrt {\dfrac{{2h}}{g}} \]
Now, for our case, as AB=BC=CD, if AB=h, then AC=2h and AD=3h. Thus,
The time taken by the body to travel the distance AB is \[{t_{(AB)}} = \sqrt {\dfrac{{2h}}{g}} \]
The time taken by the body to travel the distance AC is \[{t_{(AC)}} = \sqrt {\dfrac{{2h}}{g}} \times \sqrt 2 \]
The time taken by the body to travel the distance AD is,
\[{t_{(AD)}} = \sqrt {\dfrac{{2h}}{g}} \times \sqrt 3 \]
The time taken by the body to travel the distance BC is,
\[{t_{(AC)}} - {t_{(AB)}} = {t_{(BC)}} \\
\Rightarrow{t_{(AC)}} - {t_{(AB)}}= \sqrt {\dfrac{{2h}}{g}} \times (\sqrt 2 - 1)\]
The time taken by the body to travel the distance CD is
\[{t_{(AD)}} - {t_{(AC)}} = {t_{(CD)}} \\
\therefore{t_{(AD)}} - {t_{(AC)}} = \sqrt {\dfrac{{2h}}{g}} \times (\sqrt 3 - \sqrt 2 - 1)\]
Thus the ratio of the time taken for mentioned cases would be \[1:(\sqrt 2 - 1):(\sqrt 3 - \sqrt 2 - 1)\]

Hence option D is the correct answer.

Note: the second equation of motion can be used only because of the fact that there is constant acceleration due to gravity. If the acceleration was not constant, we cannot apply the equation as the jerk term will come into the equation, where jerk is the rate of change of acceleration. Also, here we neglect the air resistance.