The only condition for free fall as you said is that the motion of the body should be only under the influence of gravity alone. There should not be any effect of other forces like air resistance, viscous drag etc. The condition depends on the property of the material under free fall. For example, if the body has a certain mass as well as charged, it causes a deflection due to Earth's magnetic field. Also the Coriolis effect due to Earth's rotation (when dropped from space to earth) comes into play when the object falls from a much higher altitude.

For example, it's the air resistance that causes a phenomenon called terminal velocity. But this happens if the object is coming from a much higher altitude. At certain point during the free fall, the downward force of gravity on the object is balanced by the force of air resistance and then there will be no net force and the object falls with a constant velocity. This is why we are not get killed when rail drops fall on our head. Rain drops are under free fall. they start from a height of about $15km$. If there are no air resistance the velocity with which it reaches our head will be about $542 m/s$. That's a speed of $1951 km/hr$. Our head will be scattered apart.

So the presence of any other forces has a considerable impact on a body under free fall, especially when dropped from a much higher altitude.

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

Unfortunately, you are messing up quite a bit. d/g gives 2.653 s${}^2$.

You should re-read about free fall in any basic physics book. There, you'll see that when you have constant acceleration (in this case, its $g$), the height as a function of time is $$y(t) = y_0+v_0\,t+1/2\,g\,t^2$$ where $y_0$ and $v_0$ are your starting height and speed, and $y(t)$ is the height at instant $t$. You can solve for the final time knowing that the initial speed is zero, and the distance the block falls is $H=y(t_f)-y_0$. The final expression is also in any book.

The final speed can also be calculated from the equation for speed: $$v(t) = v_0 + g\,t$$

For the energy question, I could give you the formula, but I rather suggest that you first understand clearly the basic concepts of kinematics and dynamics before going to the next level of abstraction that is mechanical energy.