3(5-2x)=-4(cx+4)
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Let be the cat's speed just as it leaves the edge of the table. Then taking the point 1.3 m below the edge of the table to be the origin, the cat's horizontal position at time
is given by
and its height is
where is 9.8 m/s^2, the magnitude of the acceleration due to gravity.
The time it takes for the cat to hit the ground is with
(Unfortunately, this doesn't match any of the given options...)
The cat lands 0.75 m away (horizontally) from the edge of the table, so that its speed was
(Again, not one of the answer choices...)
I'm guessing there's either a typo in the question or answers.