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. a

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3 months ago

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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. a

t 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?

Physics

2 answers:

Keith_Richards [3.2K] · Answer 1 · 2026-02-27T04:51:28+03:00

The height from which the drops fall is 3.57m

Assuming the drops fall at a rate of 1 drop every t seconds. The initial drop takes 5t seconds to hit the ground. The second drop reaches the bottom of the 1.00 m window in 4t seconds, while the third drop reaches the top of the window in 3t seconds.

Calculate the distances covered by the second and third drops s₂ and s₃, which begin from rest at the building's roof.

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

The height of the window s is described 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 has reached the base after taking 5t seconds.

The roof height h represents the distance covered by the first drop and is expressed as,

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

the height of the roof remains at 3.57 m

Keith_Richards [3.2K] · Answer 2 · 2026-02-27T04:47:45+03:00

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

Further explanation:

As the drops descend from the roof's edge at a consistent rate, the flow rate of the drops remains constant over time. Each drop takes the same time to come down from the edge. When a drop nears the edge, the surface tension counteracts the drop's weight.

When the force of surface tension can no longer sustain the weight of the drop, it will fall from the roof.

Given:

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

Concept:

The drops descend from the roof's edge at a rate of 1 drop for every $t\text{ sec}$ .

This rate of 1 drop every $t\text{ sec}$ implies that after every $t\text{ sec}$ , a drop drops from the roof's edge.

The first drop takes 4t to reach the ground.

It takes the second drop 3t to arrive at the base of the window.

The third drop requires 2t to get to the upper edge of the window.

While the drops are airborne, they are subject to gravitational force, causing each drop to accelerate $g$ downwards.

The second equation of motion states:

$s=ut+\frac{1}{2}a{t^2}$

For free fall conditions:

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

A negative sign is used for h since the drop moves in the downward direction along the y-axis.

$h=\frac{1}{2}g{t^2}$ …… (1)

For the second drop.

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

For the third drop.

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

The difference between $h_2$ and $h_3$ will equal the window's height $1\text{ m}$ .

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

Substituting the values gives:.

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

By simplifying the expression for ${t^2}$ , we get:

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

For the 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 indicate the height of the roof above the ground.

Hence, 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

2. A ball falling under gravitational acceleration

3. Conservation of energy

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

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