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.)

Question

Nathalia Tanner

Physics

7 September, 23:45

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.)

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Answers (1)

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Shyla Larson · Answer 1

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