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
![x(t)=v_0t](https://img.qammunity.org/2020/formulas/mathematics/high-school/7bkqptw7gax7sa129ay87drvao3nb84ot4.png)
and its height is
![y(t)=1.3\,\mathrm m-gt^2](https://img.qammunity.org/2020/formulas/mathematics/high-school/jzfme3ah8lgbxz87e78l2z5hlmo5pfgjmx.png)
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
![0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\frac{1.3\,\rm m}g}\approx0.36\,\mathrm s](https://img.qammunity.org/2020/formulas/mathematics/high-school/l4ikpc6t1qns6j06ggb5plsfxt1soya2kc.png)
(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
![0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08(\rm m)/(\rm s)](https://img.qammunity.org/2020/formulas/mathematics/high-school/xb3yoi20h6qtplhom51sh5wxkla9e21chr.png)
(Again, not one of the answer choices...)
I'm guessing there's either a typo in the question or answers.