Question

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 sec, and the top-to-bottom height of the window is 2.00 m. How high above the top of the window does the flowerpot go?

Answer 1

Answer:

2.342m

Explanation:

Given

Time = 0.5 s

Height of Window = 2m

Because the pot was in view for a total of 0.5 seconds, we can assume that it took the cat 0.25 seconds to go from the bottom of the window to the top

Using this equation of motion

S = ut - ½gt²

Where s = 2

u = initial velocity = ?

t = 0.25

g = 9.8

So, we have.

2 = u * 0.25 - ½ * 9.8 * 0.25²

2 = 0.25u - 0.30625

2 + 0.30625 = 0.25u

2.30625 = 0.25u

u = 2.30625/0.25

u = 9.225 m/s ------------ the speed at the bottom of the pot

Using

v² = u² + 2gs to calculate the height above the window

Where v = final velocity = 0

u = 9.225

g = 9.8

S = height above the window

So, we have

0² = 9.225² - 2 * 9.8 * s

0 = 85.100625 - 19.6s

-85.100625 = -19.6s

S = -85.100625/19.6

S = 4.342

If 4.342m is the height above the window and the window is 2m high

Then 4.342 - 2 is the distance above the window

4.342 - 2 = 2.342m