91Ó°ÊÓ

The vertical speed of the rock is$22.0m/s$

The height of building is 30.0 m

Here, we use Newton’s third law of linear motion such that,

$V_{yf}^2=V_{yi}^2+2g(y_f-y_i)$

Where$V_{yf},V_{yi}$are the final and initial speed of the small rock and $y_f,y_i$are the final and initial positions of the rock.

The building is 30m from the street, we consider the roof of the building to be the origin $\left(y_i=0\text{m}\right).$

Thus, the street will be at with respect to roof, therefore $y_f=-30\text{m}$

Substituting these values in the above expression, we get

$\begin{array}{rcl}{V}_{yf}^2& =& 484m^2/s^2+2×9.8×30m^2/s^2\\ & =& 1072m^2/s^2\\ & =& 32.7m/s\end{array}$

Thus, the final speed of rock is $V_{yf}=32.7m/s$

Step 3: (b) Calculation of time for the whole trip

From the Newton’s first law, we know that,

$V_{yf}=V_{yi}+gt$

Here, the $V_{yf}$will have a negative sign, because it’s going downwards, i..e, the negative y-axis.

Then substituting values in above expression will give,

$\begin{array}{rcl}-32.7m/s& =& 22m/s+\left(-9.8m/s^2\right)×t\\ t& =& \frac{-54.7m/s}{-9.8m/s^2}\\ t& =& 5.6s\end{array}$

Thus the time taken in the whole trip will be double of the falling time, and that will be $2t=11.2s$.