Question

A cat drops from a shelf 4.2 ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 cm.

C) Calculate his acceleration (assumed constant) while he is stopping. Express your answer in SI units.

D) Calculate his acceleration (assumed constant) while he is
stopping, in *g*’s.

Answer #1

4.2 feet = 1.28016 m

Cat's initial velocity at 4.2 ft, u = 0

Consider cat's final velocity when it touches the ground as v

Acceleration = g = 9.8 m/s^{2}.

Distance traveled, s = 1.28016 m

Using the formula v^{2} - u^{2} = 2as,

v^{2} - 0 = 2 x 9.8 x 1.28016

v^{2} = 25.091136

v = SQRT[25.091136]

= 5.01 m/s

The cat hit the floor with a velocity v and it stopped with a
distance of 0.12 m

Final velocity, w = 0

Distance, h = 0.12 m

Consider the acceleration as a.

Using the formula, w^{2} - v^{2} = 2ah,

a = v^{2}/2h

= [25.09/(2 x 0.12)]

= **104.5464 m/s ^{2}.**

= 104.5464/9.8 g

=

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