Question

A body is dropped from a height h under acceleration due to gravity g. if \[{t_1}\] and \[{t_2}\] are the time intervals for its fall for the first one third distance and the rest $\dfrac{2}{3}$ distance, then the relation between ${t_1}$ and ${t_2}$ is ?

Answer 1

Hint:For this question the equations of motion can be used. There are three equations of motion, the equation used here is $s\;\; = \dfrac{1}{2}g{t^2}$ , which shows the relation between the distance travelled (s) and the time taken (t). Once released the only acceleration on the body will be the acceleration due to gravity. The acceleration on the body will be acting downwards.

Complete step by step answer:
Let the distance travelled by the particle in \[{t_1}\] second be d and the distance travelled by the particle in \[{t_2}\] second be\[2d\] .
Then , \[{t_1} + {\text{ }}{t_2} = {\text{ }}3d\]
From the equations of motion, Distance travelled\[ = \dfrac{1}{2}g{t^2}\]
Therefore, \[d = \dfrac{1}{2}g{t_1}^2 - - - - - \left( 1 \right)\]
and \[3d = \dfrac{1}{2}g{({t_1} + {t_2})^2} - - - - - \left( 2 \right)\]
Dividing equation 2 by equation 1, we get,
\[ \Rightarrow \dfrac{{3d = \dfrac{1}{2}g{{({t_1} + {t_2})}^2}}}{{d = \dfrac{1}{2}g{t_1}^2}}\]
Solving this equation we get,
$ \Rightarrow 3 = \dfrac{{{{({t_1} + {t_2})}^2}}}{{{t_1}^2}}$
Taking roots on both sides we get ,
$ \Rightarrow \sqrt 3 = \dfrac{{{t_1} + {t_2}}}{{{t_1}}}$
Cross multiplying the terms,
$ \Rightarrow {t_1} + {t_2} = 1.73{t_1}$
$ \Rightarrow {t_2} = (1.73 - 1){t_1}$
This is the answer to the question.
The answer can be also represented as $\dfrac{{{t_2}}}{{{t_1}}} = 0.73$ .

Note:Any body released from a height on the surface of the earth will fall down to earth with an acceleration known as the acceleration due to gravity. The value of acceleration due to gravity is $9.8m/{\sec ^2}$ . The velocity of the body will increase as time passes. This means that the velocity of the body will be maximum just before it hits the ground.
The other equations of motion are $v = u + at$ and ${v^2} - {u^2} = 2as$.
Where $v$ is the distance travelled
$u$ is the initial velocity
$a$ is the acceleration
$t$ is the time taken