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

Hint: 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.