Question

A body dropped from the top of a tower covers a distance 9x in the last second of its journey where x is the distance covered in the first second. How much time does it take to reach the ground?
A. 3sec
B. 4sec
C. 5sec
D. 6sec

Answer 1

Hint: As the body is mentioned in the question as being dropped, you could take the initial velocity as zero. Use Newton’s equation to find the distance covered in the first second. Now, recall the expression for distance covered in nth second and solve for n by substituting the distance covered as 9x and hence find the answer.

Formula Used:
Newton’s equation of motion,
$S(t)=ut+\dfrac{1}{2}a{{t}^{2}}$
Expression for distance travelled in ${n^{th}}$ second,
${{S}_{n}}=u+\dfrac{a}{2}(2n-1)$

Complete step-by-step answer:
We are given a body dropped from the top of a tower and the distance that the body travels in the first and last second is given as x and 9x respectively. What we are asked to find here is the time taken by the body to reach the ground.
Since the body is dropped, the initial velocity (u) is zero. That is,
$u=0$ …………….. (1)
From Newton’s equations of motion we have the distance covered at time t as,
$S(t)=ut+\dfrac{1}{2}a{{t}^{2}}$ …………… (2)
Since the body is falling under gravity, the acceleration here is that due to gravity (g).
So the distance covered at the first second is given by,
$S(t=1)=x=\left( 0\times 1 \right)+\dfrac{1}{2}g{{(1)}^{2}}$
Since distance covered in the first second is given as x in the question, and initial velocity is zero from (1). So,
$x=\dfrac{1}{2}g$ …………….. (3)
Let us assume that the body falls for n seconds.
Also, recall the expression for the distance travelled by a body at the ${n^{th}}$ second.
Distance travelled by the body at ${n^{th}}$ second is given as,
${{S}_{n}}=u+\dfrac{a}{2}(2n-1)$ ………………. (4)
Where,
u =initial velocity (here, u=0)
a= acceleration (here, a=g)
We are given the distance travelled by the body at nth second as 9x in the question. By substituting all the given values equation (4) becomes,
$9x=(0)+\dfrac{g}{2}(2n-1)$
$9x=\dfrac{g}{2}\left( 2n-1 \right)$ ……………. (5)
Substituting value for x from (3) in (5), we get,
$9\left( \dfrac{g}{2} \right)=\dfrac{g}{2}\left( 2n-1 \right)$
$2n-1=9$
$n=\dfrac{10}{2}=5$
Therefore, the time taken by the body to reach the ground is 5seconds.
Hence, the answer to this question is option C.

Note: The questions related to kinematics, though short, will contain all necessary quantities within the question. So, if you know how to find them, kinematics problems can be easily solved by simply substituting values. But for that, you have to read between the lines of the question and also should have strong knowledge in concepts.