x = x_o + v_x ( t – t_o )

y = y_o + v_oy ( t – t_o ) + ½ g ( t – t_o )²

v_y = v_oy+ g ( t – t_o )

v_x and v_oy are not set out data. Instead we get the actual speed, Vo, with which the shoe goes away, and the firing angle. Good, that's enough...

v_x = v_o cos 53° = 15 m/s . 0,6 = 9 m/s

v_oy = v_o sen 53° = 15 m/s . 0,8 = 12 m/s

Those are the values I put in the scheme, in the ballon that is talking about the 0 point. With all those constants, then, we setup the equations that describe the motion of the shoe.

x = 9 m/s . t₁

y = 1,25 m + 12 m/s . t – 5 m/s² . t²

v_y = 12 m/s – 10 m/s² . t

I feared this: it was a system of many equations as unknowns (4x4), in which the unknowns, if we know to interpret them, are the ones that the problem´s statement asks. It means that here finished the physics of the problem... the rest is algebra. And it is quite simple. If you you look fondly you realize that the first two are a little system (2x2) and the last two, another. They are solved separately. Look:

From the equation [1] you clear t1; it is enough if you pass 9 m/s dividing the first term. You have left t₁ = 2 s. That is the instant in which the shoe goes right above the cat (or hits him, we still don´t know). Start doing it with me, don´t be a bum.

Now this result, t₁ = 2 s, we put it into the equation [2]. Write it again, but every time it says t₁, you write 2 s. Be careful, because (2 s)² = 4 s². After replacing, you multiply, addition, subtraction and you get y₁ which gives: y₁ = 5,25 m.

But the statement was asking how much above the cat, the shoe passed, and the cat was 2 m high. Let´s call h to the height we are looking for, so

h = 5,25 m – 2 m

Now we are going to work with the other two equations, the [3] and the [4]. Always start with the equation that has fewer unknowns, in this case the [4], whose only unknown is t₂, the instant in which the shoe touches the floor. There is a problem: it can't be clear just like that, but don´t worry... the greeks invented a tiny formula that solves the problem and that you need to have always with you:

x_1.2 represents any variable, in our equation is t2, but as in any quadratic we will have two results, one of them is the one we are looking for, and the other one is useless. To use the greeks´s formula beware that the equation is equal to zero. Fortunately in our case: c is the constant term, b is the linear coefficient, and a is the quadratic coefficient:

c = 1,25 m,              b = 12 m/s,               a = – 5 m/s²

Using for once "+" and once again "-", we find the two values of t₂ that gave us:

t₂ = 2,5 s                and                 t_2´= – 0,1 s

No doubt, that one of them is good and the other is not. Now we put this result in the equation [3] and turns out

x₂ = 9 m/s . 2,5 s = 22,5 m

Then, the distance from the wall is

d = 22,5 m – 18 m

CHALLENGE: The cat didn´t see the shoe coming out the window and, therefore, he do not know from what height left. Moreover, he thinks that the shoe is a flying object of oblique shots fired from the floor. Without operations than mentally, say how much should the cat assume that lasted the flight of the shoe.