What you are doing wrong is using equations that apply *only* when the acceleration is constant to a situation where the acceleration is variable.

If you had a function that gave the velocity *vs* time, you could integrate that from $t_0$ to $t_{final}$.

EDIT:
For example. suppose the velocity, $v$, as a function of time is given by:$$v=18-12t+0.1t^2$$Then the displacement, $d$, at a time $T$, is given by:$$d=\int_{0}^{T}{v}dt=\int_{0}^{T}{18-12t+0.1t^2}dt=18T-6T^2+\frac{0.1}{3}T^3$$

Given a graph, one solution is to plot the curve *very* carefully on some graph paper, and then count the squares between the velocity curve and the x-axis (the time axis). Remember that squares below the x-axis are negative...

This necro-bumped question deserves an answer that touches the delicate matter of finding the right solution when many roots exist.

In most cases, the "wrong" solutions can still be understood what they would hypothetically mean, even if they have no physical realization.

Let's say the time gap (last second) is $t_0$ and the displacement ratio to the total height is $\eta =z/h$ (to avoid explicit numbers).

We have

$$h=\frac12 gt^2;\quad \eta h=\frac12 g(t-t_0)^2$$

Subtracting these, we obtain
$$h(1-\eta)=\frac12 g(2tt_0-t_0^2)$$

Plug it back into the first equation:

$$\frac12 g(2tt_0-t_0^2)\frac{1}{1-\eta}=\frac12 gt^2$$
Cancel and rearrange:
$$(1-\eta)t^2-2tt_0+t_0^2=0$$
Solve:
$$t=t_0\frac{1\pm \sqrt\eta}{1-\eta}=t_0\frac{1}{1\mp\sqrt{\eta}}$$

Your numbers $\eta=0.48$ and $t_0=1\,\rm s$ give solutions $t=\{3.3{\,\rm s},0.6{\,\rm s}\}$

Here, we have to discard the second solution, as it would measure the height between times 0.6s and -0.4s, which was **before** the fall started (where it would have been, if it wasn't just dropped from the top but thrown upwards, and we just started the clock at the highest position).

If we plug both solutions into the first height equation $\frac12 gt^2$, we get

$$\color{green}{ h(3.3{\,\rm s})=52\,\rm m},\quad \color{red}{h(0.6{\,\rm s})=1.7\,\rm m}$$

Notice how the first solution (computed by the original poster) is **correct**, as the second one corresponds to the one where the travel time is shorter than the measurement interval.

## Best Answer

They are pretty much the same thing. If they are not, then then one is just a constant offset of the other.

When we talk about the position of an object, we usually express that position in terms of a position vector $\mathbf r(t)$. This vector can be expressed based on a coordinate system, for example in Cartesian coordinates $$\mathbf r(t)=x(t)\hat x+y(t)\hat y+z(t)\hat z$$

Note that $\mathbf r(t)=0$ at the origin, i.e. when $x=y=z=0$. Additionally, you could draw the path this vector traces in time, and it would be seeing the position of the object at various times points. For your case $x(t)=z(t)=0$, and so you can easily plot $y(t)$ vs. $t$.

Displacement $\Delta\mathbf r(t)$ is the change in position of an object relative to some initial position $\mathbf r_0$. i.e. $$\Delta\mathbf r(t)=\mathbf r(t)-\mathbf r_0$$

Note that displacement is $0$ when $\mathbf r(t)=\mathbf r_0$. Typically, we just choose $\mathbf r_0=0$ so that displacement and position are equal. However, you could instead use the displacement from some other position that is not the origin. Then displacement and position would not be the same thing, but they would just be offset by the constant position vector $\mathbf r_0$. Another way to look at this is that you are explicitly stating a shift in your coordinates so that the origin is at $\mathbf r_0$ compared to the original coordinate system.

With all of this in mind, physicists usually use position and displacement interchangeably, and they might not be as precise as what is discussed in this answer. However, what is actually meant is typically obvious from the context. I would say it is a distinction that you should not really worry about.