Charles Cisneros

2021-11-23

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?

Opeance1951

Beginner2021-11-24Added 26 answers

We have the time the pot takes to fall a distance $y-y-0=1.9$ mm and $g=9.8$ , from this we can calculated the velocity of the pot when it just reached the top of the windows in question. Take the positive y direction to be downward

$y-{y}_{0}={v}_{0}t+\frac{1}{2}{>}^{2}$

$v}_{0}=\frac{y-{y}_{0}-\frac{1}{2}{>}^{2}}{t$

$v}_{0}=\frac{\left(1.9\right)-(4.9\times {0.38}^{2})}{0.38}=3.138\text{}\frac{m}{s$

The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this infirmation we can get the distance between the sill and the top of the window below.

${v}^{2}={v}_{0}^{2}+2g(y-{y}_{0})$

$y-{y}_{0}=\frac{{v}^{2}-{v}_{0}}{2g}=\frac{{3.138}^{2}-0}{2\times 9.8}=0.5$ m

Result: The top of the windows is 0.5 m below the sill of the upper window

The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this infirmation we can get the distance between the sill and the top of the window below.

Result: The top of the windows is 0.5 m below the sill of the upper window

22+64

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