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 0.47 s, and the top-to-bottom height of the window is 2.14 m. How high above the window top does the flowerpot go?

Answer 1

Time while going up = time while going down.
One-way time = 0.47 / 2 = 0.235 s
While going up,
s = ut + 1/2 at²; where u is the initial velocity upwards and a is the acceleration due to gravity downwards.
2.14 = u(0.235) + 1/2 x -9.81 x (0.235)²
u = 10.2 m/s

Max height attained can be obtained by:
2as = v² - u²; where v is 0
s = -(10.2)² / -9.81 x 2
s = 5.30 m
The height above the window is:
5.30 - 2.14
= 3.16 m