Question

A body is falling from height h it takes ${t_1}$ time to reach the ground. The time taken to cover a first half of height is
A. ${t_2} = \dfrac{{{t_1}}}{{\sqrt 2 }}$
B. ${t_2} = \sqrt 2 {t_1}$
C. ${t_2} = \sqrt 3 {t_1}$
D. None of these

Answer 1

Use the kinematic equation to express the distance covered by the body. The initial velocity of the body is zero when it is dropped from a certain height. Express the kinematic equation to reach the first half of the total distance and solve the two equations to get the time taken for the body to cover the first half.

Formula used:
Kinematic equation, $d = ut + \dfrac{1}{2}a{t^2}$,
where, d is the distance, u is the initial velocity, t is the time and a is the acceleration.

Complete step by step answer:
We can express the total distance covered by the body using kinematic equation as follows,
$h = u{t_1} + \dfrac{1}{2}gt_1^2$
Here, h is the total height, $u$ is the initial velocity of the body, g is the acceleration due to gravity and ${t_1}$ is the time required to reach the ground.
Since the body is dropped from height h, its initial velocity must be zero. Therefore, we can write the above equation as,
$h = \dfrac{1}{2}gt_1^2$ …… (1)
Now, let us express the kinematic equation to reach the first half of the total distance as follows,
$\dfrac{h}{2} = \dfrac{1}{2}gt_2^2$
$\Rightarrow h = gt_2^2$ …… (2)
Here, ${t_2}$ is the time taken to cover the first half the total height.
Equating equation (1) and equation (2), we get,
$\dfrac{1}{2}gt_1^2 = gt_2^2$
$\therefore {t_2} = \dfrac{{{t_1}}}{{\sqrt 2 }}$

So, the correct answer is option A.

Note: While using kinematic equations in vertical motion, the acceleration of the particle should be replaced by the acceleration due to gravity. Also, for the vertical downward motion, all the terms in the kinematic equation have the negative sign and therefore, we can neglect the negative sign for each term. For the vertical upward motion, only the acceleration due to gravity should have the negative sign.