Question

The ratio of displacements covered by a freely falling body under gravity of $g = 10m{s^{ - 2}}$ in the last two successive seconds is $7:9$. Then height from which body is dropped is given as
A. $45m$
B. $80m$
C. $125m$
D. $180m$

Answer 1

In the above question, we are asked to find the height of the body which is freely falling. For that we are provided with the ratio of the displacements of the two successive seconds. We can consider the time to be $t,t - 1,t - 2$ respectively. then applying the equation of motion, we can find our respective answer.

Complete step by step solution:
Let us consider the time starts at $t$then successive seconds are taken as $t - 1$ and $t - 2$. The distances covered in these seconds are ${s_1},{s_2}$ and ${s_3}$ respectively. Now using the equation for the freely falling body which is $v = u + at$.
where v is the final velocity of the object, u is the initial velocity of the object and a be the acceleration of the object.
Forming the equation of motion for all the successive seconds by taking ${v_1},{u_1}$ as final and initial velocities for ${s_1}$ and vice versa.
${v_1} = 0 + g\left( {t - 2} \right)$ as initial velocity is zero.
$\Rightarrow{v_2} = g\left( {t - 1} \right)$
$\Rightarrow{v_3} = gt$
Now using equation ${v^2} - {u^2} = 2as$ where s is the displacement of the object
In these situations, for $t - 2$ final velocity be given by ${v_1}$ and substituting in these equations:
${v_1}^2 - {v_2}^2 = 2g{s_1}$
Using the above values
${g^2}{\left( {t - 1} \right)^2} - {g^2}{\left( {t - 2} \right)^2} = 2g{s_1}$ $\ldots \ldots \left( 1 \right)$
Similarly,
{v_3}^2 - {v_2}^2 = 2g{s_2} \\\ \Rightarrow {g^2}{t^2} - {g^2}{\left( {t - 1} \right)^2} = 2g{s_2} \\\ $\ldots \ldots \left( 2 \right)$
And we given the ratio of displacements $\dfrac{{{s_1}}}{{{s_2}}} = \dfrac{7}{9}$ $\ldots \ldots \left( 3 \right)$
Equating the three marked equations
$\dfrac{{{{\left( {t - 1} \right)}^2} - {{\left( {t - 2} \right)}^2}}}{{{t^2} - {{\left( {t - 1} \right)}^2}}} = \dfrac{7}{9}$
After solving, we get
$\dfrac{{2t - 3}}{{2t - 1}} = \dfrac{7}{9} \\\ \Rightarrow 18t - 27 = 14t - 7 \\\ \Rightarrow 4t = 20 \\\ \Rightarrow t = 5\sec \\\$
But in the question, we are asked for the height from which the body is dropped. hence, using this equation $s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow s = 0t + \dfrac{1}{2}\left( {10} \right){\left( 5 \right)^2} \\\ \therefore s= 125m$
Hence, the height with which the body is dropped is $125m$

The correct option is C.

Note: For a freely falling body we can take acceleration to be acceleration of gravity i.e., $g = 10m{s^{ - 2}}$ which is acting downwards and can take it to be positive. The initial velocity of the freely falling body is taken to be zero. The final velocity of the body falling upward is also taken as zero.