Which Step Does It Land On?

We observe that the $x$ coordinate of the projectile can be expressed as such: $x=12\cos{75^\circ}t\\t=\frac{x+4}{12\cos{75^\circ}}$

Then, we observe that the $y$ coordinate of the projectile can be expressed as such: $y=12\sin{75^\circ}t-\frac{g}{2}t^2$

Substituting the first equation into the second, we find that $y=\frac{-g}{288\cos^2{75^\circ}}x^2+\left(\tan{75^\circ}-\frac{8g}{288\cos^2{75^\circ}}\right)x-\frac{g}{18\cos^275^\circ}$

Then, by observing the graph, we can deduce that the answer is 2.765. One may use desmos for this.

Relevant wiki: 1D Kinematics Problem Solving

I assume that everyone knows how to solve 2D kinematic equations, so here's a rough outline of the solution.

1) Based on the x velocity, determine the two times (tx) at which the projectile passes over the x boundaries of each step.

2) Based on the initial y velocity and the influence of gravity, determine the second instance in time (ty) at which the projectile y position matches the altitude of each step. The first instance in time corresponds to a negative x value, and is thus useless.

3) If the solution to (2) is between the two numbers from (1), we have a candidate intersection.

4) Find the highest (in terms of y position) step with a candidate intersection.

5) Find the projectile x coordinate corresponding to the time of intersection with the staircase step from (4)

The solution spreadsheet and projectile plot are given below:

I used desmos. We know that the equation of a projectile is given by $: y = x\tan \theta(1 - \frac {x}{R})$ . You can find a proof here . Also, the range ( $R$ ) of this projectile is $\frac {v_0^{2}\sin(2\theta)}{g} = \frac {72}{g}$ . Graphing this at the given values gives us a path that looks like -

Just by eyeballing it, we can see that $x \approx 2.7$ .(you can draw the steps if you like).