Q74P - University Physics with Modern Physics

Question

A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell?

Step-by-Step Solution

Answer

The top if the window will below from flowerpot fell is 0.5 m. 

Step1: Identification of given data

Height = 1.90 m.

Time = 0.380 s

Step 2: Calculation for the distance

We have the time the pot takes to fall a distance  $y-y_0=1.9mm$

 And  $g=9.8$

Take the positive y direction to be downward

$\begin{array}{r}y-y_0=v_{0}t+{\left(\frac{1}{2}\right)}^2 \\ v_0=\frac{y-y_0-{\left(\frac{1}{2}\right)}^2}{t} \\ v_0=\frac{\left(1.9\right)-\left(4.9\times {0.38}^2\right)}{0.38}\\ v_0=\frac{y-y_0-{\left(\frac{1}{2}\right)}^2}{t} \\ v_0=3.138\mathrm{m}/\mathrm{s} \end{array}$ 

The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this information.

We can get the distance between the sill and the top of the window below.

$\begin{array}{r}{v}^2={v_0}^2+2g\left(y-y_0\right)\\ y-y_0=\frac{v^2-v_0}{2g} \\ y-y_0=\frac{{3.138}^2-0}{2\times 9.8} \\ y-y_0=0.5\mathrm{m} \end{array}$