Question

Let A, B, C, D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the time interval of descent through AB, BC and CD are in the ratio:

• A. $1:\sqrt{3}-\sqrt{2}:\sqrt{3}+\sqrt{2}$
• B. $1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}$
• C. $1:\sqrt{2}-1:\sqrt{3}$
• D. $1:\sqrt{2}:\sqrt{3}-1$

Answer 1

The correct option is B $1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}$

We have a standard expression for time of descent when a body is released from rest. $t=\sqrt{\frac{2h}{g}}$ ..........(i)
Let $AB=h⟹AC=2h\text{and}AD=3h$
Time for descent up to B from A: $t_{AB}=\sqrt{\frac{2h}{g}}$
Time for descent up to C from A: $t_{AC}=\sqrt{\frac{4h}{g}}=\sqrt{\frac{2h}{g}}\times \sqrt{2}$
Time for descent up to D from A: $t_{AD}=\sqrt{\frac{6h}{g}}=\sqrt{\frac{2h}{g}}\times \sqrt{3}$
Time taken along BC: $t_{BC}=t_{AC}-t_{AB}=\sqrt{\frac{2h}{g}}\times (\sqrt{2}-1)$
Time taken along CD: $t_{CD}=t_{AD}-t_{AC}=\sqrt{\frac{2h}{g}}\times (\sqrt{3}-\sqrt{2})$
Hence the ratio $t_{AB}:t_{BC}:t_{CD}=1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}$
Note: Equation (i) can be derived from the laws of motion with uniform acceleration
Consider $s=ut+\frac{1}{2}a{t}^2$.

When a body is released from rest from a height $h$ and acceleration due to gravity is $g$, we have $s=h;a=g;$ $t$ being the time of flight which gives $h=\frac{1}{2}g{t}^2$