What strategy should I use to approach this problem? If my straight line approximation doesn't work [closed]

Question

recently i wrote an assignment in one of the questions i got two different answers for the same question by considering a straight line motion

The question was somewhat like this

THIS IS A HOMEWORK QUESTION AND I AM NOT ASKING THE EXACT ANSWER I JUST WANT GUIDANCE

A particle is thrown obliquely with a velocity of $15\sqrt{5}$ $metres$ $per$ $second$ and 45 degrees to the horizontal. $100$ meters from the point of projection starts stairs in which each step is $1$ meter in height and $1$ meter in width. Keeping this in consideration, which step should the particle hit?

I calculated the horizontal range of this projectile from the formula $$R=\frac{u^2\sin(2\theta)}g=112.5\,m$$

prescribed value of $g$ is $10$ metres per second square

That means the particle would have to hit any one of the steps.

Proceeding forwards with the formula of $$x \tan\theta\left(1-\frac{x}R\right)=y$$ where $x$ is the horizontal displacement of the particle and $y$ is the vertical displacement.

Through which I can calculate its height, $y=11.111$, and as we know that the remaining horizontal distance to be covered by the particle before coming to the level of point of projection is $112.5-100=12.5$ by constructing a right angled triangle like this

I know that particle won't travel a straight path in the influence of gravity, but as the distance is not too far from it's final point we can consider its path to be straight for some time.

as the point will be at the midpoint of rectangle and making the rectangle into four triangles

By Pythagorean theorem, we get the hypotenuse in every triangle as $8.35 meters$ then the point where they meet is approximately 8.35 meters on the line of stairs.

By converting this value into steps i.e. as we know one step measures $\sqrt{2}$ then for 8.32 it would be $5.883128419$ $meters$. Therefore, it would hit the wall of fifth step.

But would this approximation be correct as we considered a straight line motion of the particle so would it hit 5th step or 6th step

how can I look at the problem and others like it so that I can imagine and formulate a solution on my own?

i just have an idea of pythagorean theorem tosolve this and nothing else

Answer 1

Using g = 10 m/s, and the provided information, I get the following:

This means the projectile would strike between 6 and 7 meters and end up on the step that is 6 meters from the ground.

Equations:
For $x_i$, I select the values between 100 and 112 inclusive. For y, the equation is
$$y_i=x_i\cdot tan(\theta )\cdot \left (1-\frac {x_i}{x_{max}}\right )$$
Using 112.5 m for $x_{max}$.

Answer 2

One potential way of going about finding which step you hit is the following. You can parameterize the motion of the particle in the plane by $(x(t), y(t))$, where $t$ is time and $x$ and $y$ are as follows,

$$y(t) = y_i + (v_i)_yt - \frac{g}{2}t^2, \quad x(t) = x_i + (v_i)_x t.$$

Fortunately we can kill the parameter $t$, via the substitution $t = \frac{x - x_i}{(v_i)_x}$ into the left equation above. Doing so yields (taking $x_i = y_i = 0$):

$$y = (v_i)_y \frac{x}{(v_i)_x} - \frac{g}{2}\left(\frac{x}{(v_i)_x}\right)^2.$$

Because of symmetry, $(v_i)_y/(v_i)_x = 1$. Thus, we have

$$ y = x - \frac{g}{2(v_i)_x^2}x^2,\tag{1}$$

Where we've written the height $y$ as a quadratic function of the horizontal distance $x$.

Now consider the line touching the top of the stairs and the line touching the bottom of the stairs (like the one in your picture) given by the equations

$$ y = x - 99 \quad \text{and} \tag{2}$$ $$ y = x - 100,\tag{3}$$

respectively. Now by solving for $x_1$ such that $(1) = (2)$ (taking the positive solution) and solving for $x_2$ such that $(1) = (3)$ (again taking the positive solution) you can deduce which step you hit. That is if there exists $n$ such that $n \le x_1 < x_2 \le n + 1$, then you hit the $(n - 99)$th step, and if there exists $n$ such that

$$n-1 < x_1 < n < x_2 < n+1$$

then you hit the vertical piece after the $(n - 100)$th step. Also, notice that since $\frac{dy}{dx} = 1 - \frac{g}{(v_i)_x^2}x = 0$ for $x = \frac{(v_i)_x^2}{g} = (15\sqrt{5/2})^2/9.8 \approx 57m$, we know that after roughly $57m$ the particle is "falling" and thus we must have one of the above two cases.