Question

A stone is dropped from the top of a tall tower after $1$ second another stone is dropped from a balcony $20 m$ below the top. If both the stones reach the ground at the same instant, then the height of the tower is:
A. $51.25 m$
B. $21.25 m$
C. $31.25 m$
D. $3.125 m$

Answer 1

If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. Use the equation of motion in both the conditions and subtract the equation to obtain the value of time and then substitute it in the equation.

Formula used:
Equation of motion, $s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is displacement, $u$ is initial velocity, $t$ is time, $a$ is acceleration.

Complete step by step answer:
Let us consider the height of the tower to be ${h_t}$.
From the given question, we know that the height of the balcony is ${h_b} = {h_t} - 20\;{\rm{m}}$ and the time difference when the both stones are thrown is $\Delta t = 1\;{\rm{s}}$.
The time difference when the both stones are thrown is,

$$\Delta t = {t_1} - {t_2}\\\ \Rightarrow 1 = {t_1} - {t_2}\\\ \Rightarrow {t_2} = {t_1} - 1$$

We know that when both the stones are dropped, their initial velocity is equal to zero, $u = 0$.
Since the stone is moving downwards the acceleration due to gravity will be taken as $g = + 10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

Caste I: When the stone is dropped from the tower:
Let us use the equation of motion,
$s = ut + \dfrac{1}{2}g{t^2}$
Substitute the expression in the above equation, we have,
$\left( {0 - {h_t}} \right) = 0 + \dfrac{1}{2}gt_1^2$
${h_t} = \dfrac{1}{2}gt_1^2$ ... (I)

Caste II: When the stone is dropped from the balcony:
Similarly, from the equation of motion,
${h_b} = 0 + \dfrac{1}{2}gt_2^2$
$\left( {{h_t} - 20} \right) = 0 + \dfrac{1}{2}g{\left( {{t_1} - 1} \right)^2}$... (II)

Now we subtract equation (I) and (II),

$$20 = \dfrac{1}{2}gt_1^2 - \dfrac{1}{2}g{\left( {{t_1} - 1} \right)^2}\\\ \Rightarrow 20 = \dfrac{1}{2}g\left( {t_1^2 - {{\left( {{t_1} - 1} \right)}^2}} \right)\\\ \Rightarrow 40 = 10 \times \left( {2t - 1} \right)$$

Simplifying the above equation,
$40 = 10\left( {2{t_1} - 1} \right)\\\ \Rightarrow{t_1} = 2.5\;{\rm{s}}$
Now substitute the above value in equation (I), we get,

$${h_t} = \dfrac{1}{2} \times 10 \times {\left( {2.5} \right)^2}\\\ \therefore{h_t} = 31.25\;{\rm{m}}$$

Thus, the height of the tower is 31.25 m and option (C) is correct.

Note: Make sure the height of the balcony is taken into account its height is 20 m below the height of the tower. Do not use 20 m as the height of the balcony. Someone has the stone at the top of the tower to drop, so the initial velocity is taken as zero in this type of question. If you are considering the height of the tower or building as positive, then the datum is its stop.