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

Answer:

1.54 m

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

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.84-\frac{1}{2}\times -9.81\times 0.27^2}{0.27}\\\Rightarrow u=8.14\ m/s$

$v=u+at\\\Rightarrow v=8.14-9.81\times 0.27\\\Rightarrow v=5.5\ m/s$

This will be the initial velocity while going up higher

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-5.5^2}{2\times -9.81}\\\Rightarrow s=1.54\ m$

The the flower pot will go 1.54 m above the window