Question

During the last second of its free fall, a body covers half of the total distance travelled. Calculate:
i) the duration of the fall
ii) the approximate height from which the body falls.

Answer 1

We can use the second equation of motion to calculate the required value of time and height from which the body falls. When the body is falling, it will experience acceleration due to gravity g whose value is $9.8m/{s^2}$
Equation of motion to be used:
$s = ut + \dfrac{1}{2}a{t^2}$ where, s is distance covered, u is initial velocity, a is acceleration and t is time taken.

Complete Step by step answer: It is given in the question, last half of the total distance is covered by the body in one second. Let the total time taken by t. the diagram of the fall can be given as:

Using the following equation of motion, we can calculate the required values:
$s = ut + \dfrac{1}{2}a{t^2}$ ____________ (1)
The total distance covered will be the total height (h) in time (t) under the force of gravity is given as:
$h = ut + \dfrac{1}{2}g{t^2}$
But the initial velocity of object is zero, so it becomes:
$h = \dfrac{1}{2}g{t^2}$ _________ (2)
Now, it is given that
Half the distance (height) is covered in the last second, so the first half of the distance will be covered in total time minus one second. The object does not have any initial velocity and as it is falling down, the acceleration it suffers is due to gravity. Substituting these in (1), we get:
$\dfrac{h}{2} = (0)t + \dfrac{1}{2}g{\left( {t - 1} \right)^2}$ ________ (3)
i) To calculate the value of t. we will substitute the value of h from (2):
$\dfrac{1}{2} \times \dfrac{1}{2}g{t^2} = + \dfrac{1}{2}g{\left( {t - 1} \right)^2} \\\ \Rightarrow \dfrac{1}{2} \times \dfrac{1}{2}g{t^2} = + \dfrac{1}{2}g\left( {{t^2} + 1 - 2t} \Rightarrow \right)\left( {\because {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right) \\\ \Rightarrow {t^2} = 2{t^2} + 2 - 4t \\\ \Rightarrow {t^2} - 4t + 2 = 0 \\\$
This is a quadratic equation in t and can be factorized to get its value using:
$t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here,
b = -4, a = 1 and c = 2, substituting:
$\Rightarrow t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ \Rightarrow t = \dfrac{{ - ( - 4) \pm \sqrt {{{(4)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}} \\\ \Rightarrow t = \dfrac{{4 \pm \sqrt {16 - 8} }}{2} \\\ \Rightarrow t = \dfrac{{4 \pm 2\sqrt 2 }}{2} \\\ \Rightarrow t = 2 \pm \sqrt 2 \left( {\because 2 = common} \right) \\\$
We have two values of t:
$t = 2 - \sqrt 2 s = 0.59s \\\ \Rightarrow t = 2 + \sqrt 2 s = 3.41s \\\$
The value of t should be greater than 1, because it covers half the distance in 1 second, so it will be covering the complete distance in more than 1 second. So we will ignore the first and consider the second value of t.
Therefore, the duration of the fall of the body is 3.41 seconds.
ii) To find the value of height, we will substitute the calculated value of t in (2):
$h = \dfrac{1}{2} \times 9.8 \times {(3.41)^2}\left( {\because g = 9.8m/{s^2}} \right) \\\ \Rightarrow h = 56.9 \approx 57 \\\$
Therefore, the approximate height from which the body falls is 57m.

Note: when a body falls freely, it is always under the force of gravity and thus experiences acceleration due to gravity. If for a falling object, it is not given that it was in motion, we consider that it was at rest before the fall and take its initial velocity zero.
A quadratic equation is that where the maximum power of the variable is 2. For any quadratic equation $a{x^2} + bx + c = 0$ , the factors can be given as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
We use identities to simplify the equation and obtain the value of variable, the identity used here is:
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$