Question

A particle is dropped from the top of a high tower. The ratio of time in falling successive distances h is:
A) $1:(\sqrt 2 - 1):(\sqrt 3 - \sqrt 2 )......$
B) $1:3:5$
C) $1:2:3$
D) $2:3:5$

Answer 1

In this question we have to find the ratio of time in falling successive distances h. For this we are going to use the formula of height or distance covered from dropped from a height. Using this formula we will find the ratio of time.

Complete step by step solution:
Given,
The displacements are successive, so if the particle is travelling h distance in time ${t_1}$ then after time ${t_1} + {t_2}$ the particle will travel h+h distance and after time ${t_1} + {t_2} + {t_3}$the distance travelled by the particle will be h+h+h.
Formula used,
$\Rightarrow h = \dfrac{1}{2}g{t^2}$
After time ${t_1}$
$\Rightarrow h = \dfrac{1}{2}g{t_1}^2$
$\Rightarrow {t_1} = \sqrt {\dfrac{{2h}}{g}}$…. (1)
Displacement after time ${t_1} + {t_2}$
$\Rightarrow h + h = \dfrac{1}{2}g{({t_1} + {t_2})^2}$
$\Rightarrow 2h = \dfrac{1}{2}g{({t_1} + {t_2})^2}$
$\Rightarrow {t_1} + {t_2} = \sqrt {\dfrac{{4h}}{g}}$
$\Rightarrow {t_2} = \sqrt {\dfrac{{4h}}{g}} - {t_1}$
Putting the value of ${t_1}$ from equation (1)
$\Rightarrow {t_2} = \sqrt {\dfrac{{4h}}{g}} - \sqrt {\dfrac{{2h}}{g}}$…. (2)
Displacement after time ${t_1} + {t_2} + {t_3}$
$\Rightarrow h + h + h = \dfrac{1}{2}g{({t_1} + {t_2} + {t_3})^2}$
$\Rightarrow 3h = \dfrac{1}{2}g{({t_1} + {t_2} + {t_3})^2}$
$\Rightarrow {t_1} + {t_2} + {t_3} = \sqrt {\dfrac{{6h}}{g}}$
$\Rightarrow {t_3} = \sqrt {\dfrac{{6h}}{g}} - {t_1} - {t_2}$
Putting the values of ${t_2}$ and ${t_3}$
$\Rightarrow {t_3} = \sqrt {\dfrac{{6h}}{g}} - \sqrt {\dfrac{{2h}}{g}} - \sqrt {\dfrac{{4h}}{g}} + \sqrt {\dfrac{{2h}}{g}}$
$\Rightarrow {t_3} = \sqrt {\dfrac{{6h}}{g}} - \sqrt {\dfrac{{4h}}{g}}$….. (3)
Now, the ratio of time${t_1}$, ${t_2}$, ${t_3}$and so on….
$\Rightarrow {t_1}:{t_2}:{t_3}:...... = \sqrt {\dfrac{{2h}}{g}} :\left( {\sqrt {\dfrac{{4h}}{g}} - \sqrt {\dfrac{{2h}}{g}} } \right):\left( {\sqrt {\dfrac{{6h}}{g}} - \sqrt {\dfrac{{4h}}{g}} } \right):......$
$\Rightarrow {t_1}:{t_2}:{t_3}:...... = 1:(\sqrt 2 - 1):(\sqrt 3 - \sqrt 2 ):......$

Hence from above calculation we see that the ratio of time in falling from successive distances h is ${t_1}:{t_2}:{t_3}:...... = 1:(\sqrt 2 - 1):(\sqrt 3 - \sqrt 2 ):......$.

Note: In this question we have used the formula of height for falling particles. To solve such types of questions we should know this formula. While doing calculation of this type of questions we should be very careful as the calculation is sort of difficult. As we have seen that in this question the distance is successive so we should know what successive distance means.
In the same case if the distance is successive will time also be successive or not one should be careful about that. As in this question the successive distance covered in time t will continue till the particle touches the ground. So the ratio of time will continue till that point.