Question

A body initially at rest falls and covers half of the total distance in the last second of its fall. If the acceleration due to gravity, $g = 9.8{\text{m/}}{{\text{s}}^2}$. Find the total time taken to fall in seconds.
A) $2{\text{s}}$
B) $\left( {2 + \sqrt 2 } \right){\text{s}}$
C) $\left( {4 - \sqrt 2 } \right){\text{s}}$
D) $3{\text{s}}$

Answer 1

The body falling from its rest position can be considered to be falling with the acceleration due to gravity. As the body is initially at rest, its initial velocity will be zero. Newton’s equation of motion, under constant acceleration, relating the displacement of the body, its initial velocity, its acceleration and time taken can be used to find the time of fall.

Formula Used: Newton’s laws of motion, under constant acceleration, depicting the distance covered by a body is given by, $s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the total distance covered by the body, $u$ is its initial velocity, $t$ is the time taken to cover the distance and $a$ is the acceleration of the body.

Complete step by step answer:
Step 1: List the parameters provided in the question.
The body is initially at rest. This implies that the initial velocity of the body, $u = 0$ .
Let $s$ be the height of the fall or the total distance covered as the body falls and let $t$ be the time taken to cover the distance $s$ .
Step 2: Express the total distance of the body using Newton’s equation of motion.
Newton’s equation of motion depicting the distance covered by a body is given by,
$s = ut + \dfrac{1}{2}a{t^2}$ -------- (1)
where $s$ is the total distance covered by the body, $u$ is its initial velocity, $t$ is the time taken to cover the distance and $a$ is the acceleration of the body.
Here, the acceleration of the body is $g$ and the initial velocity $u = 0$ .
Substituting for $g = 9.8{\text{m/}}{{\text{s}}^2}$ and $u = 0$ in equation (1) we get, $s = \dfrac{1}{2}9.8{t^2}$
Thus the total distance covered by the body is given by $s = \dfrac{1}{2}9.8{t^2}$ ------- (2).
Step 3: Express the distance covered by the body in the last second of its fall using Newton’s equation of motion.
Given, the distance covered in the last second is half of its total distance i.e., $s$ will be $\dfrac{s}{2}$ .
Now, if $t$ is the total time of the fall then the last second of the fall will be $t - 1$ .
Then using equation (1), the distance covered in the last second can be expressed as $\dfrac{s}{2} = \dfrac{1}{2}9.8{\left( {t - 1} \right)^2}$ or we get, $s = 9.8{\left( {t - 1} \right)^2}$ ------- (3).
Step 4: Using equations (2) and (3) find the time taken for the fall.
Equation (2) gives $s = \dfrac{1}{2}9.8{t^2}$ and equation (3) gives $s = 9.8{\left( {t - 1} \right)^2}$
Divide equation (3) by (2) we get, $\dfrac{s}{s} = \dfrac{{2 \times 9.8{{\left( {t - 1} \right)}^2}}}{{9.8{t^2}}}$
Cancel the similar terms in the numerator and denominator to get, $\dfrac{{2{{\left( {t - 1} \right)}^2}}}{{{t^2}}} = 1$
This can be reduced by cross multiplying and expanding the term ${\left( {t - 1} \right)^2}$
Then we have $2{t^2} - 4t + 2 = {t^2}$ or ${t^2} - 4t + 2 = 0$ ------ (4)
We can solve equation (4) using the quadratic formula given by, $t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
From equation (4), we have the coefficients of $t$ as $a = 1$ , $b = - 4$ and $c = 2$
Substituting the values for $a = 1{\text{, }}b = - 4{\text{, }}c = 2$ in the above formula we get, $t = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - \left( {4 \times 1 \times 2} \right)} }}{{2 \times 1}}$
Simplifying we get, $t = \dfrac{{4 \pm \sqrt {16 - 8} }}{2} = 2 \pm \sqrt 2$
$\therefore$ the total time taken for the fall is $t = 2 + \sqrt 2 {\text{ s}}$ . Hence, the correct option is b.

Note: A quadratic equation represented by, $a{x^2} + bx + c = 0$ is solved using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Here the equation (4) given by, ${t^2} - 4t + 2 = 0$ is quadratic in $t$ . Thus we can solve for $t$ in (4) using the same quadratic formula. Here both values of $t$ ( $t = 2 + \sqrt 2$ and $t = 2 - \sqrt 2$ ) are positive.