Assuming non-relativistic velocities, the power radiated by a charge accelerating at constant acceleration $a$ is given by the Larmor formula:
$$P = \frac{e^2 a^2}{6\pi \epsilon_0 c^3} $$
To do the calculation properly is surprisingly complicated, but it's easy show that the effect of the radiation on the electrons fall is negligible. If the electron falls a distance $h$ then the time it takes is given by:
$$ h = \frac{1}{2}gt^2 $$
so:
$$ t = \sqrt{\frac{2h}{g}} $$
If we assume the electron is accelerating at a constant rate of $g$, the total energy radiated is just power times time or:
$$ E_{rad} = \frac{e^2 g^2}{6\pi \epsilon_0 c^3} \sqrt{\frac{2h}{g}} $$
In your question $h$ is 1000m, so:
$$ E_{rad} = 7.83 \times 10^{-51}J $$
The potential energy change is, as you say, just $mgh$:
$$ E_{pot} = m_e g h = 8.94 \times 10^{-27} J $$
So the ratio of the radiated energy to the potential energy is about $10^{-24}$, and therefore the effect of the radiation on the electron's fall is entirely negligible.
Response to comment:
The power radiated from the electron produces a force that opposes the acceleration due to gravity. Assume we can ignore the deviations from accelerating at a constant rate $g$, then in a small time $dt$ the energy radiated is $Pdt$. The energy is force times distance ($dx$) so to get the force we divide by the distance:
$$ F = P\frac{dt}{dx} = \frac{P}{v} = \frac{P}{\sqrt{2gh}} $$
using $v^2 = 2as$. The acceleration produced by this force is just $F/m_e$, so the net acceleration on the electron is:
$$ a_{net} = g - \frac{P}{m_e \sqrt{2gh}} $$
So the electron does accelerate slightly more slowly than $g$, but the difference between the acceleration and $g$ is inversely proportional to distance fallen so it gets increasingly negligible the further the electron falls.
You've probably spotted that the above equation says the force should be infinite at the moment you release the particle. That's because as you approach the moment of release it's no longer safe to make the approximation that you can ignore the change in the acceleration due to radiation.
Buoyant force is a force acts exactly opposite to gravitational force.
The slower velocity of ball moving thru liquid is due to drag of viscous fluid.
When we say weightlessness of ball, it only means there is no force acting on the mass externally. it always has mass, and due to density difference(Liquid and ball) it should fall downwards.
If the drag force is very high (High viscous), then you can see that ball moves upwards in the liquid when you jump down. The reason is...because of the terminal velocity difference between you and the ball inside the liquid. You hit terminal velocity only after the ball hits terminal velocity because you both are moving in 2 different medium. Your velocity will be very high compared to the ball which starts moving towards the cap.
This could be calculated by following known values
Specific gravity of liquid,
Material/Density of ball.
I tried an experiment to find out whether my reasoning is right by using 2 different liquid with 1.29 specific gravity liquid and 1.35 specific gravity liquid. I filled it in a bottle and placed the cap on after dropping a steel ball in it. Now shake the bottle(Vertically) to same distance and mark how far the ball moves. Do the experiment with high viscous liquid and mark. You will find the difference.
Best Answer
Right before impact, there is no acceleration, such that forces by air resistance and gravity cancel. Right after impact, the velocity has changed sign (no energy loss), such that also the air resistance changes sign. Instead of $a=g-g$ you get $a=-g-g=-2g$.