Question

A body dropped from the top of the tower covers a distance $7x$ 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) $3s$
B) $4s$
C) $5s$
D) $6s$

Answer 1

Hint: Free fall occurs where the only force acting on an object is gravity. Hence acceleration on earth is constant, the distance of an object falls is directly proportional to the time spent falling. If we throw a ball in upward direction , then it will come back because when the ball is going in upwards direction, its speed will be less as compared to when it comes down. This is because of acceleration and also it is produced due to the force of gravity.

Formula Used:
We will be using the formula of distance travelled. The distance fallen after a time t is given by the formula $d = \dfrac{1}{2}g{t^2}$.

Complete step by step solution:
The distance of a free falling object has fallen from a position of rest is also dependent upon the time of fall. The distance travelled is defined as how much path an object has covered to reach its destination in a given period.
$d = \dfrac{1}{2}g{t^2}$
Where,
g is the acceleration due to gravity ($9.8m{s^{ - 2}}$)
d is the distance the object travelled in meters,
t is the time in seconds.
Therefore distance covered in ${1^{st}}$ second is equal to $\dfrac{1}{2}g$
We know that distance travelled in ${n^{th}}$ second is given as $d = \dfrac{1}{2}g(2n - 1)$
Let it fall for n second $n = 1$
Therefore, $7x = 0 + \dfrac{g}{2}(2n - 1)$
$ \Rightarrow 7 \times \dfrac{g}{2} = \dfrac{g}{2}(2n - 1)$
Simplifying we get,
$ \Rightarrow 2n = 8$
$ \therefore n = 4s$
Thus the total time taken to fall is $4$ seconds.
Answer is Option (B), $4s$.

Note: The free falling objects are in a state of acceleration. They are accelerating at a rate of $9.8m{s^{ - 2}}$. The velocity of a free falling object is changing by $9.8m{s^{ - 1}}$. The S.I unit of acceleration is meter per second squared. The value of g depends on the location.