A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

Question

3 years ago

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A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

. how fast is the flower pot traveling when it hits the street? (give your answer in ft/s and in mi/hr, given that the acceleration due to gravity is 32 ft/s2 and 1 ft/s = 15/22 mi/hr.)

Physics

1 answer:

Harlamova29_29 [7] · Answer 1 · 2022-04-12T16:58:58+03:00

H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer: 133.9 ft/s or 91.3 mi/h