# Pushing a ball up an inclined slope; how far up from the ground did the ball rise?

Imagine you have a slope with an angle $\theta$ relative to the ground. We know from doing many experiments of flicking a wooden ball up the ramp that the furthest distance it gets displaced in the $x$ direction is $0.444$ m in $4.8$ s. After that, it turns back and rolls down the incline. Assume that $a_y$ is $9.8$ $m/s^2$. How far up from the ground did the ball rise?

I'm getting ridiculous numbers here:

First, I want to find the initial velocity. At the furthest point the ball goes (a.k.a. $0.444$ m. $x$ - direction displacement), $v_{fy}$, or final velocity should be $0$, right? If I want to find its initial velocity, I would do the following:

\begin{aligned}0-v_{iy}&=-9.81\cdot4.8\\ v_{iy}&=47.088\quad\mathrm{m/s}\\ v_{fy}^2&=v_{iy}^2+2(-9.8)y\\ y&=113.1\quad\mathrm{m}\end{aligned}

What's wrong with my understanding here? Am I not using certain pieces of information I should have used?

It seems fine to me, but I'm not an expert on these problems. I think the reason you're getting such a big number is because this ramp isn't very wide, only $44\,\mathrm{cm}$ at its base. But the ball manages to keep rolling for almost $5\,\mathrm{s}$ and, if you've ever flicked a ball up a ramp, that's a rather long time. This would suggest you've got a pretty steep ramp. Also, presumably, this is a frictionless ramp, which might not give you a realistic answer.
Furthermore, your answer boils down to $y=9.81\frac{(\Delta t)^2}{2}$, which may be big when $\Delta t=4.8\,\mathrm{s}$ but with a more realistic $\Delta t=1\,\mathrm{s}$, you'd have $y\approx 5\,\mathrm{m}$, which is also more realistic.
You are not using at all that the ball is moving on an inclined plane. What it means is that the ball is acted on not only by gravity, but by the normal force on the plane. If you draw a diagram, you can see that the gravity has a component along the plane equal to $a=g\sin\theta$. The ball is launched with a velocity $v$ (unknown for now) along the plane. It slows down uniformly, so at $t=4.8$s it comes to rest. You can get from here that $$v=g\sin\theta t$$ The motion along the plane is such that after time $t$ the ball went a distance $l$: $$l=vt-\frac{1}{2}g\sin\theta t^2=\frac{1}{2}g\sin\theta t^2$$ The projection of $l$ on the horizontal axis is $x$ and on the vertical axis is $y$. We have $x=l\cos\theta$ and $y=l\sin\theta$. By multiplying the above equation by $l$ we get $$l^2=\frac{1}{2}gyt^2$$ From the definition of $x$ and $y$ we also get $$x^2+y^2=l^2=\frac{1}{2}gyt^2$$ or $$y^2-\frac{1}{2}gyt^2+x^2=0$$ The only unknown in this equation is $y$. You should get two solutions. One of them is large, so the motion is almost vertically up (assumes a high initial speed), and one very small (very shallow plane, and slow initial velocity)