Interesting problem regarding free falling body?

Question

A dog sees a flowerpot sail up and then back past a window 1.1 m high. If the total time the pot is in sight is 0.54 s, find the height above the top of the window to which the pot rises.

Please explain the steps.

Answer 1

Each time the pot passes the windows takes 0.54/2 = 0.27 seconds. This is so because the trajectory is symmetric.

Consider when the flowerpot is moving downward.

Using:

g = acceleration of gravity = 9.8 m/s^2

t = time flowerpot reaches the top of the window

T = time flowerpot reaches the bottom of the window

A standard equation for something having constant acceleration is that the distance it travels in time "t" is 1/2 the acceleration times t^2. So:

s1 = (1/2)gt^2  .... distance pot has traveled to top of window

s2 = (1/2)gT^2 .... distance pot has traveled to bottom of window

s2 - s1 = height of window = 1.1

T - t = 0.27 seconds so T = 0.27 + t

s2 - s1 = (g/2)(T^2 - t^2)

1.1 = 4.9[(0.27 + t)^2 - t^2]

1.1 = 4.9[0.0729 + 0.54t]

0.2245 = 0.0729 + 0.54t

t = (0.2245 - 0.0729)/0.54

t = 0.2807 seconds

So the flowerpot has fallen for 0.2807 seconds before it reaches the top of the window.

s = (1/2)gt^2 = 4.9(0.2807)^2

s = 0.396 meters

Note: if they meant that each time the pot passes the window it takes 0.54 seconds then my answer must be corrected. But the problem did state "total time" so I took this to mean the time it was visible both going up and coming down.