A cat drops from a shelf 3.6 ft above the floor and lands on all four feet. His

Question

A cat drops from a shelf 3.6 ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 cm.

Part A:
Calculate his speed when he first touches the floor (ignore air resistance). Express your answer to two significant figures and include the appropriate units.

Part B:
Calculate how long it takes him to stop. Express your answer to two significant figures and include the appropriate units.

Part C:
Calculate his acceleration (assumed constant) while he is stopping, in m/s^2. Express your answer to two significant figures and include the appropriate units.

Part D:
Calculate his acceleration (assumed constant) while he is stopping, in g's. Express your answer to two significant figures.
Calculate

Explanation / Answer

Shelf height h = 3.6 ft                       = 3.6 x 0.304 m                       = 1.0944 m Stopping distance S = 12 cm                                 = 0.12 m (A).   His speed when he first touches the floor u = [ 2gh ]                                                                           = 4.6 m / s   Since g = 9.8 m / s 2 ( B) Time taken to stop t = ? From the relation v 2 - u 2 = 2aS Accleration a = [v 2 - u 2 ] / 2S Where v = final velocity                = 0 Substitue values we get a = 89 m / s 2 So, required time t = ( v - u ) / a                               = 5.1 x 10 -2 s (C).His acceleration a = 89 m / s 2 ( D).His acceleration a =  [ (89 m / s 2 )/ (9.8 m / s 2 ) ] g                                    = 9.1 g

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