Question

A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally in possesion of a high-speed, high-precision timing system, notices that it takes 0.18 s to fall past his 3.5 m tall window. How far above the top of the window is the ledge from which the pot fell? m

Answer 1

Answer:

the window is 21m below the point where the pot fell

Explanation:

The person has made a measurement of the time and distance that the pot travels, so we can use the free decayed distance equation

       Y = Vo t - ½ g t²

  With the distance equal to the height of the window and the measured time of 0.18 s, so we can calculate the initial speed with which it is at the top point of the window

      Vo = (Y + ½ g t2) / t

      Vo = (3.5 + ½ 9.8 0.18²) /0.18

      Vo = 20.3 m / s

Fearing the speed with which the flowerpot begins its passage through the window we calculate the distance it takes to reach this speed.

     

     Vf² = Vo² - 2g Y

 

As the flowerpot falls from rest, the initial speed is zero and the final speed is the speed with which it reaches the edge of the window Vf = 20.3 m / s

      Vf² = - 2 g Y

Let's clarify in these problems the positive coordinate axis is up

      Y = -Vf² / 2 g

      Y = -20.3² / 2  9.8

      Y = -21 m

This means that the window is 21m below the point where the pot fell