Question

A drowsy cat is looking at a window from across the room, and sees a flowerpot that sail first up and then down past the window. The pot is in view for a total of 0.54 s, and the top-to-bottom height of the window is 1.84 m. How high above the window top does the flowerpot go?

Answer 1

1.54 m

Step-by-step explanation:

The time the cat sees the pot going up will be half of the total time the cat saw the pot. The same is true for the pot going down.

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`s=ut+(1)/(2)at^2\\\Rightarrow u=(s-(1)/(2)at^2)/(t)\\\Rightarrow u=(1.84-(1)/(2)* -9.81* 0.27^2)/(0.27)\\\Rightarrow u=8.14\ m/s`

`v=u+at\\\Rightarrow v=8.14-9.81* 0.27\\\Rightarrow v=5.5\ m/s`

This will be the initial velocity while going up higher

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-5.5^2)/(2* -9.81)\\\Rightarrow s=1.54\ m`

The the flower pot will go 1.54 m above the window