Question

Drops of water fall from the roof of a building 18 m high in regular intervals of time. When the first drop reaches the ground at the same instant fourth drop starts to fall. What are the distances of the second and third drops from the roof?
A. 6 m and 2 m
B. 6 m and 3 m
C. 8 m and 2 m
D. 4 m and 2 m

Answer 1

Assume the interval of time between the two drops as t and then determine the time of flight for the first drop. Use the kinematic equation to determine the time interval between the falls of two drops. Identify the time interval for the second and third drop and use the kinematic equation to determine the distance from the roof.

Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$
Here, s is the displacement, u is the initial velocity, g is the acceleration due to gravity and t is the time.

Complete step by step answer:
We assume the interval of fall between the two drops is t. Since we have three intervals between four drops, the interval of fall between first drop and fourth drop will be 3t. Thus, the time of flight for the first drop is 3t. Let’s use kinematic equation to express the distance travelled by the first drop as follows,
$s = u{t_1} + \dfrac{1}{2}gt_1^2$
Here, s is the height of the roof from the ground, u is the initial velocity of the drop, ${t_1}$ is the time of flight for the first drop and g is the acceleration due to gravity.

Since the drops are falling freely from the roof, their initial velocities are zero. Therefore, we can write the above equation as,
$s = \dfrac{1}{2}gt_1^2$
Substituting 18 m for s, $10\,{\text{m/}}{{\text{s}}^2}$ for g and 3t for ${t_1}$, we get,
$18 = \dfrac{1}{2}\left( {10} \right){\left( {3t} \right)^2}$
$\Rightarrow 3.6 = 9{t^2}$
$\Rightarrow {t^2} = 0.4$
$\Rightarrow t = 0.63\,{\text{sec}}$
Thus, we get the time interval between the falls of two drops is 0.63 seconds.

Now, for the second drop, the time of flight is 2t. Using the kinematic equation, we can calculate the distance travelled by the second drop from the roof as follows,
${s_2} = \dfrac{1}{2}g{\left( {2t} \right)^2}$
$\Rightarrow {s_2} = 2g{t^2}$
Substituting $10\,{\text{m/}}{{\text{s}}^2}$ for g and 0.63 s for t in the above equation, we get,
${s_2} = 2\left( {10} \right){\left( {0.63} \right)^2}$
$\Rightarrow {s_2} = 8\,{\text{m}}$
Therefore, the distance of the second drop from the roof is 8 m.

Now, for the third drop, the time of flight is t. Using the kinematic equation, we can calculate the distance travelled by the third drop from the roof as follows,
${s_3} = \dfrac{1}{2}g{t^2}$
Substituting $10\,{\text{m/}}{{\text{s}}^2}$ for g and 0.63 s for t in the above equation, we get,
${s_2} = \dfrac{1}{2}\left( {10} \right){\left( {0.63} \right)^2}$
$\therefore {s_2} = 2\,{\text{m}}$
Therefore, the distance of the second drop from the roof is 2 m.

So, the correct answer is option C.

Note: The crucial step of this solution is to determine the time of flight of the drops. Remember, there are three intervals between the four drops and not four intervals. Also, while using a kinematic equation for the downward motion of the drop, you can take the positive sign for both distance and acceleration due to gravity as both heads towards the ground.