Let $v_0$ 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 $t$ is given by

$x(t)=v_0t$

and its height is

$y(t)=1.3\,\mathrm m-gt^2$

where $g$ 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 $t$ with

$0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\dfrac{1.3\,\rm m}g}\approx0.36\,\mathrm s$

(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 $v_0$ was

$0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08\dfrac{\rm m}{\rm s}$

(Again, not one of the answer choices...)

I'm guessing there's either a typo in the question or answers.

6 0

3 years ago

If a number cube is labeled with the numbers 1 through 6. what is the probablity of rolling a 1 or a 5 on the first roll?

kow [346]

It would be 2/6 or 1/3.

4 0

4 years ago

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