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?

Answer 1

Answer:

Distance =0.5 m

Explanation:

We have given time the pot takes to fall a distance

y-y₀=1.9m

And Gravity g=9.8m/s²

From this we can calculate the velocity of the pot when

it just reached the top of the window.take positive y direction to be downward

$y-y_{o}=v_{o}t+(1/2)gt^{2} \\ v_{o}=\frac{y-y_{o} -1/2gt^{2} }{t}\\ v_{o}=\frac{(1.9)-(4.9*0.38^{2} )}{0.38}\\ v_{o}=3.138m/s$

The flowerpot falls from rest and we have velocity it gain until it reaches the top of the window below,From this information we can find distance between the sill and the top of the window as:

$v^{2}=(v_{o})^{2}+2g(y-y_{o} )\\ (y-y_{o} )=\frac{v^{2}-v_{o}}{2g}\\ (y-y_{o} )=\frac{(3.138)^{2}-0}{2(9.8)}\\ (y-y_{o} )=0.5m$