A cat chases a mouse across a 1.1 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 1.6 m from...

Hello!

Denote the initial speed of the cat as `V_0.` The direction of the initial velocity is horizontal (assuming the table is horizontal). Denote the height of the table as `H_0` and the given distance (horizontal) as `D_0.`  

After sliding, the cat participates in two movements which we can consider separately. The first is the horizontal one with the constant speed `V_0` (there is no force to change this speed). The second is the...

Hello!

Denote the initial speed of the cat as `V_0.` The direction of the initial velocity is horizontal (assuming the table is horizontal). Denote the height of the table as `H_0` and the given distance (horizontal) as `D_0.`  

After sliding, the cat participates in two movements which we can consider separately. The first is the horizontal one with the constant speed `V_0` (there is no force to change this speed). The second is the vertical one (downwards) under the gravity force.

So the horizontal distance from the table is `D(t) = V_0 t,` the height is `H(t) = H_0 - (g t^2)/2.` From the first equation we can find the time `t_1 = D_0/V_0.`  The cat hits the floor at this moment, so `H(t_1)` must be zero. This gives the equation `H_0-(g t_1^2)/2=0,` or `H_0=g/2 (D_0/V_0)^2.`

Solve this equation for `V_0:`  `D_0/V_0 = sqrt((2 H_0)/g),` or `V_0 = D_0/sqrt((2 H_0)/g).`  In numbers it is about 3.38 (m/s). This is the answer.

Note: if the given distance of `1.6 m` is not a horizontal one but is the distance between the edge points of a cat's trajectory, then `D_0=sqrt(1.6^2-1.1^2) approx 1.16(m)` by the Pythagorean Theorem.