Question

Water droplets fall from a tap on to the floor $5.0m$ below at regular intervals of time. The first drop strikes the floor when the fifth drop begins to fall. The height at which the third drop will be from the ground, at the instant when the first drop strikes the ground is (Take $g=10m{{s}^{-2}}$)
$A)\text{ }1.25m$
$B)\text{ 2}.15m$
$C)\text{ 2}.75m$
$D)\text{ 3}.75m$

Answer 1

Hint: This problem can be solved by using the equations of motion for constant acceleration using which we will find out the time between successive drops and hence, find out the displacement of the third drop at the time when the first drop reaches the ground.
Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
We will use the equations of motion for constant acceleration to find out the time gap between the drops falling from the tap in terms of the time required for a drop to fall and reach the ground. Using this, we can find out the displacement covered by the third drop when the first drop reaches the ground.
The displacement $s$ covered by an object in time $t$ moving in uniform motion with constant acceleration $a$ is given by
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --(1)
Where $u$ is the initial velocity of the body.
Now, let us analyze the question.
Let the time taken by a drop to fall to the ground be $T$.
The height of the tap from the ground is $h=5m$.
When a drop is falling, it is under uniform motion with constant acceleration $a=g=10m{{s}^{-2}}$.
Let the time gap between two successive drops falling from the tap be $t$.
The initial velocity of each drop falling from the tap will be $u=0$
According to the question, the fifth drop begins to fall when the first drop reaches the ground.
$\therefore T=4t$ --(2)
Now, using (1), we get
$5=0\left( T \right)+\dfrac{1}{2}g\left( {{T}^{2}} \right)$
$\therefore 5=\dfrac{1}{2}g{{T}^{2}}$
$\therefore 5\times 2=10{{T}^{2}}$
$\therefore 10=10{{T}^{2}}$
$\therefore {{T}^{2}}=1$
Square rooting both sides, we get
$\sqrt{{{T}^{2}}}=\sqrt{1}$
$\therefore T=1s$ --(3)
Therefore, the time required for each drop to fall to the ground is $1s$.
Now, putting (3) in (2), we get
$1=4t$
$\therefore t=\dfrac{1}{4}=0.25s$
Hence, the time gap between the falling of two successive drops from the tap is $0.25s$.
Now, let the initial time when the first drop starts to fall be ${{t}_{1}}=0s$. Therefore, the third drop starts to fall at ${{t}_{3}}={{t}_{1}}+2t=0+2\left( 0.25 \right)=0.5s$.
The first drop touches the ground at $T=1s$ and at that time, the third drop has fallen for a time of $T-{{t}_{3}}=1-0.5=0.5s$.
Therefore, using (1), the displacement $d$ of the third drop by that time will be
$d=0\left( 0.5 \right)+\dfrac{1}{2}\left( 10 \right){{\left( 0.5 \right)}^{2}}$
$\therefore d=\dfrac{1}{2}\times 10\times 0.25=1.25m$
Now, the height of the drop above the ground at this point will be nothing but the total height of the tap above the ground minus the displacement covered by the drop in the direction from the tap towards the ground.
Therefore, this height $h'$ is
$h'=h-d=5-1.25=3.75m$
Therefore, the height of the drop above the ground at the required time is $3.75m$.

Therefore, the correct option is $D)\text{ 3}.75m$.

Note:
Students can make a mistake of writing the time gap between the first drop falling and the fifth drop falling as $5t$ instead of $4t$ as written in the answer above. However, students must understand that $t$ is the time gap between two successive drops falling from the tap and hence, there will be a single $t$ time gap between the first and second drop falling, $2t$ time gap between the first and third drop falling and so on for the ${{n}^{th}}$ drop, the time gap between the ${{n}^{th}}$ drop and the first drop will be $\left( n-1 \right)t$.