Question

A particle is dropped from the top of a tower. During its motion, it covers $9/{{25}^{th}}$ part of the tower in the last one second. Find the height of the tower.

Answer 1

Hint: This problem can be solved by finding out the total time of travel of the particle by using the formula for the distance travelled by a particle in a specific second for constant acceleration motion.
Formula used:
For a body subjected to motion with constant acceleration $a$ in one dimension, the displacement ${{s}_{n}}$ travelled by the body in the ${{n}^{th}}$ second is given by
${{s}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)$
where $u$ is the initial velocity of the body.
The distance $s$ covered by a body under constant acceleration $a$ in a time $t$ is given by,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where $u$ is the initial acceleration of the body

Complete step by step answer:
As explained in the hint, we will find out the total time of the travel of the body by equating the formula for the distance travelled in a specific second by the particle with the information given in the question. Therefore let us proceed to do the above and analyze the question.
Let the total height of the tower be $h$.
The total time travelled by the particle is $t$.
The acceleration of the body is the acceleration due to gravity $g$.
We will consider the last second, that is the ${{t}^{th}}$ second.
Distance travelled by the particle in the last second is ${{s}_{t}}=\dfrac{9}{25}h$
Since, the particle is dropped and not thrown downwards with any velocity, the initial velocity of the particle is $u=0$.
For a body subjected to motion with constant acceleration $a$ in one dimension, the displacement ${{s}_{n}}$ travelled by the body in the ${{n}^{th}}$ second is given by
${{s}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)$ --------(1)
where $u$ is the initial velocity of the body.
Therefore, using (1) and the information we got after analyzing the question, we get,
$\dfrac{9}{25}h=0+\dfrac{1}{2}g\left( 2t-1 \right)$ --------(2)
Now, the distance $s$ covered by a body under constant acceleration $a$ in a time $t$ is given by,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --------(3)
where $u$ is the initial acceleration of the body.
The total height $h$ of the tower is the distance travelled by the particle in the time $t$. Therefore, using (3), we get,
$h=0\left( t \right)+\dfrac{1}{2}g{{t}^{2}}$
$\therefore h=\dfrac{1}{2}g{{t}^{2}}$ -------(4)
Using (4) in (2), we get,
$\dfrac{9}{25}\times \dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g\left( 2t-1 \right)$
$\therefore \dfrac{9}{25}{{t}^{2}}=\left( 2t-1 \right)$
$\therefore 9{{t}^{2}}=25\left( 2t-1 \right)=50t-25$
$\therefore 9{{t}^{2}}-50t+25=0$
$\therefore t=\dfrac{-\left( -50 \right)\pm \sqrt{{{\left( -50 \right)}^{2}}-4\left( 9 \right)\left( 25 \right)}}{2\left( 9 \right)}$ $\left( \because \text{if }a{{x}^{2}}+bx+c=0,\text{ }x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)$
$\therefore t=\dfrac{50\pm \sqrt{2500-900}}{18}=\dfrac{50\pm \sqrt{1600}}{18}=\dfrac{50\pm 40}{18}$
$\therefore t=\dfrac{50+40}{18}=\dfrac{90}{18}=5s$
or,
$t=\dfrac{50-40}{18}=\dfrac{10}{18}=0.555s<1s$
This is not possible as it is stated that in the last second of travel the particle covers a part of the height of the tower. Hence, the total time must be greater than $1s$.
$\therefore t\ne 0.555s$
$\therefore t=5s$ --(5)
Using (5) in (4), we get,
$h=\dfrac{1}{2}g\left( {{5}^{2}} \right)=\dfrac{1}{2}\left( 9.8 \right)\left( 25 \right)=122.5m$ $\left( \because g=9.8m/{{s}^{2}} \right)$
Hence, the total height of the tower is $122.5m$.

Note: This problem could also be solved by considering that the rest of the height of the tower was covered in $t-1$ seconds and using the equation of motion for constant acceleration to solve this equation. However, that would add some unnecessary variables and lengthy calculations and it is better to use the direct formula for the distance travelled in a specific second and equate it to the information given in the question.