Question

A window washer drops a brush from a
scaffold on a tall office building.What is the speed of the falling brush after
2.41 s?acceleration due to gravity is 9.8 m/s
and How far has it fallen in that time?

Answer 1

i think you may be able to use \[\Delta y=v_{0}t+\frac{ 1 }{ 2 }at^2\]

Answer 2

it was wrong

Answer 3

\[\Delta y=v_{0}t+\frac{ 1 }{ 2 }at^2\] but write it like this, \[y=y_{0}+v_{0}t+\frac{ 1 }{ 2 }at^2\] and v(0)=0 and set y(0) to zero and you just have \[y=\frac{ 1 }{ 2 }at^2\] i think that will work.

Answer 4

With the \(y\)-axis pointing downwards from the window washer

The \(y\) position of the falling brush is
\[y(t)=y_{0}+{v_y}_0t+\tfrac12a_yt^2\]
Choosing the origin to be where the window washer is \(y(0)=y_{0}=0\)
\[y(t)={v_y}_0t+\tfrac12a_yt^2\]
The velocity in the \(y\) direction
\[\dot y(t) = v_y(t)={v_y}_0+a_yt\]
Initial velocity is zero, \(v_y(0)={v_y}_0=0\)

Hence the velocity in the \(y\) direction of the falling brush at a time \(t\) is
\[v_y(t)=a_yt\]
Acceleration due to gravity is \(a_y=g = 9.8[\text{m/s}^2]\).
The time in question is \(t_1=2.41 [\text s]\).

The \(y\) position at \(t_1\) is how far the brush has fallen
\[y(t_1)=\tfrac12gt^2\]