Question

Water drops fall from the roof of a building 20 m high at regular time intervals. If the first drop strikes the floor when the sixth drop begins to fall, the height of the second and fourth drops from the ground at that instant are \[\left( {g = 10{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\]

Answer 1

Hint:This problem is based on projectile motion. In the solution, we will use the formula of time taken to transverse the path. Also, the second equation of motion is utilized to calculate the height of the second and fourth drop.

Complete step by step answer:
Given:The height of the building is h=20 m. The acceleration due to gravity is g=10 $m{s^2}$.
We know the formula of time taken by each drop to reach the ground.
$t = \sqrt {\dfrac{{2h}}{g}} $
On plugging in the values of height and gravitational acceleration in the above expression, we get,
\[
t = \sqrt {\dfrac{{2 \times 20\;{\rm{m}}}}{{10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \\
\Rightarrow t = 2\;{\rm{s}}
\]
We know the expression of interval between each drop is given by,
$
T = \dfrac{t}{5}\\
\Rightarrow T = \dfrac{{2\;{\rm{s}}}}{5}\\
\Rightarrow T = 0.4\;{\rm{s}}
$
The height of the second drop after 2 s is calculated by using the formula,
$
{h_2} = h - \dfrac{1}{2} \times g \times {\left( {t - T} \right)^2}\\
\Rightarrow{h_2} = 20 - \dfrac{1}{2} \times 10 \times {\left( {2 - 0.4} \right)^2}\;{\rm{m}}\\
\Rightarrow{h_2} = 7.2\;{\rm{m}}
$
The height of the fourth drop after 2 s is calculated by using the formula,
$
{h_4} = h - \dfrac{1}{2} \times g \times {\left( {t - 3T} \right)^2}\\
\Rightarrow{h_4} = 20 - \dfrac{1}{2} \times 10 \times {\left( {2 - 3 \times 0.4} \right)^2}\;{\rm{m}}\\
\Rightarrow {h_4} = 16.8\;{\rm{m}}
$
Now, we will calculate the height of the second and fourth drops from the ground at that instant.
$D = {h_4} - {h_2}$
On putting the values in the above distance formula, we get,
$
D = 16.8\;{\rm{m}} - 7.2\;{\rm{m}}\\
\therefore D = 9.6\;{\rm{m}}
$

Thus, the height of the second and fourth drops from the ground at that instant is $D = 9.6\;{\rm{m}}$.

Note: In such types of problems, to find out each drop's height, we will use the second equation of motion because all the parameters in the question fulfill the requirement of the equation. One of the critical points in the solution where a student will make a mistake in calculating the interval between each drop.