Question

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. at this instant, the second and third drops are exactly at the bottom and top edges of a 1.00-m-tall window. how high is the edge of the roof?

Answer 1

The height of the roof is 3.57m

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

Answer 2

The height of the edge of the roof is $\boxed{3.57\text{ m}}$ or $\boxed{357\text{ cm}}$.

Further explanation:

When drops fall from the edge of a roof at a steady rate, the rate of flow does not change with time. Each drops take same time to fall from the edge. If a drop is at the verge of fall, surface tension will balance the weight of the drop.

When surface tension reaches to its extreme value and weight of the drop exceeds the maximum value of surface tension, water drop will fall from the edge of a roof.

Given:

The height of the window is $1\text{ m}$.

Concept:

Let the rate at which drops fall from the edge of a roof is 1 drop per $t\text{ sec}$.

The rate 1 drop per $t\text{ sec}$ indicates that after every $t\text{ sec}$ one drop falls from the edge of a roof.  

The time taken by first drop to reach ground is 4t.

Time taken by second drop to reach at the bottom of the window is 3t.

Time taken by third drop to reach at the top of the window is 2t.

During the whole time when drop is in air, it is subjected to a gravitational pull. So, the acceleration of each drop will be $g$ in downward direction.

The second equation of motion is:  

$s=ut+\frac{1}{2}a{t^2}$  

For free fall.  

$\begin{aligned}u&=0 \hfill \\s&=- h \hfill \\a&=- g \hfill \\ \end{aligned}$  

Negative sign is taken for h as drop travels in the negative direction of y axis.

$h=\frac{1}{2}g{t^2}$                                       …… (1)

For second drop.

$\begin{aligned}{h_2}&=\frac{1}{2}g{\left( {4t} \right)^2} \\&=8g{t^2} \\ \end{aligned}$  

For third drop.

$\begin{aligned}{h_3}&=\frac{1}{2}g{\left( {3t} \right)^2} \\&=4.5g{t^2} \\ \end{aligned}$

 

The difference of $h_2$ and $h_3$ will be the length of the window $1\text{ m}$.

${h_2}-{h_3}=1\,{\text{m}}$

Substitute the values.

$\begin{aligned}8g{t^2} - 4.5g{t^2}&=1\,{\text{m}} \hfill \\{\text{3}}{\text{.5g}}{{\text{t}}^2}&=1\,{\text{m}} \hfill \\3.5\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right){t^2}&=1\,{\text{m}} \hfill \\ \end{aligned}$

Simplify the above expression for ${t^2}$.  

$\begin{aligned}{{\text{t}}^2}&=\frac{{1{\kern 1pt} {\text{m}}}}{{3.5\left( {9.81{\kern 1pt} {\text{m/}}{{\text{s}}^{\text{2}}}} \right)}} \\&=0.02913\,{{\text{s}}^{\text{2}}} \\ \end{aligned}$  

For first drop.

$\begin{aligned}{h_1}&=\frac{1}{2}g{\left( {5t} \right)^2} \\&=12.5\,g{t^2} \\&=\left( {12.5} \right)\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {0.02913\,{{\text{s}}^{\text{2}}}} \right) \\&=3.57\,{\text{m}} \\ \end{aligned}$  

$h_1$ will be the height of the roof from ground.

Thus, the height of the edge of the roof is $\boxed{3.57\text{ m}}$ or $\boxed{357\text{ cm}}$.

Learn More:

1.  The motion of a body under friction brainly.com/question/4033012

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Water, drops, edge, roof, steady rate, fifth, starts, fall, just, hits, ground, instant, second, third, fourth, exactly, bottom, top, 1.00 m, tall, 100 cm, 1.00 meter, 1 meter, height, window, 3.57 m, 3.57 meter, 357 cm.