Question

A ball is dropped from ${{\text{9}}^{{\text{th}}}}$ stair of a multi-storeyed building reaches the ground in ${\text{3}}$ seconds. In the first second of its free fall, it passes through ‘$n$’ stair then ‘$n$’ equal to:
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Answer 1

First of all, we will find the distance covered by the body in ${\text{3}}$ seconds and in one second separately. Then we will compare the respective distances to find the result.

Complete step by step answer:
In the given problem, we are supplied the following data:
The ball is dropped from the ${{\text{9}}^{{\text{th}}}}$ stair of a multi-storeyed building.
The ball reaches the ground in ${\text{3}}$ seconds.
We are asked to find the number of stairs it passes through in the first second.
This is a problem which is based on the free fall of objects. Initially the velocity of the ball is zero, when it was released. Let the height of a floor be $h$ .
First, we will use an equation from motion:
$S = ut + \dfrac{1}{2}g{t^2}$ …… (1)
Where,
$S$ indicates the vertical distance the ball had travelled.
$u$ indicates the initial velocity.
$t$ indicates time.
$g$ indicates acceleration due to gravity.
So, substituting the required values in the equation (1), we get:

$$S = ut + \dfrac{1}{2}g{t^2} \\\ 9h = 0 \times 3 + \dfrac{1}{2} \times g \times {3^2} \\\ h = \dfrac{g}{2} \\\$$

Now, we will find the distance covered in first second:
We can apply the equation (1) again to find the same,

$$S = ut + \dfrac{1}{2}g{t^2} \\\ S = 0 \times 3 + \dfrac{1}{2} \times g \times {1^2} \\\ S = \dfrac{g}{2} \\\$$

We can write:
$\dfrac{g}{2} = h$
So,
The distance covered in the first second is $h$ .
Now, we can write the height of $n$ floors as $nh$ to calculate how many floors have passed in one second.
Mathematically,

$$nh = h \\\ n = 1 \\\$$

Hence, in the first second of its free fall, it passes through ‘$n$’ stairs then ‘$n$’ equal to $1$ .
The correct option is A.

Note: This is a problem based on free fall. It should be remembered that when a body just falls from a height, the initial velocity is zero. The body falls fully under the action of gravity.