Question

A small rock is thrown vertically upward with a speed of22.0 m/s from the edge of the roof of a 30.0-m-tall building. Therock doesn’t hit the building on its way back down and lands onthe street below. Ignore air resistance. (a) What is the speed of therock just before it hits the street? (b) How much time elapses fromwhen the rock is thrown until it hits the street?

Short answer

(a) The speed of the rock just before it hits the street will be$V_{yf}=32.7m/s$, and

(b) Total time elapsed when the rock is thrown until it hits the street is $11.2s$.

Step-by-step solution

Given data

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

The height of building is 30.0 m

(a) The speed of rock just before it hits the street

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

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

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,

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

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