If a ball is kept on a platform which is suspended horizontally from a helicopter, what would it take for the helicopter to drop the ball? So I want to describe the premise better. By a helicopter I mean something that can stay suspended in air perfectly and can move around in any direction freely. To generalize the motion of the helicopter mathematically, assume that the position of the helicopter in each dimension is a continuous and differentiable function. You can assume that the platform is a square plate with 4 strings from each vertex that are connected to a single point at the bottom of the helicopter. Assume ideal conditions like the wind of the helicopter blades doesn't affect the platform and that there's no air resistance or friction. At t=0, the platform is perfectly horizontal and the helicopter, the ball and the platform are at rest.
So the question is, what does it take for the helicopter to drop the ball? In my opinion it's certainly harder than it seems. If it starts to accelerate uniformly and slowly in a particular direction, the platform could tilt of course, but in the frame of reference of the platform, the forces on the ball will be balanced perfectly by the pseudo force due to the acceleration.
One condition in which the ball would fall is if it the downward acceleration of the helicopter would exceed the acceleration due to gravity. In that situation the ball would "levitate" from the platform's perspective and then the helicopter can just move away horizontally and drop the ball. But I don't know if that's a complete answer.
Also, how would the answer be affected if the strings couldn't "bend"? As in they were rigid beams that could move around just as freely?
 A: There is a nice demonstration by Rhett Allain of essentially the same setup as you are discribing, but smaller. The tray that Rhett is holding up is square, and strings attached to the four corners, the four strings go up and are combined, Rhett is holding up the tray.
The title of the youtube video of Rhett's demonstration is Physics and serving drinks
If you need to walk some distance carrying a tray of drinks you need to walk quite evenly, otherwise the fluid in the various glasses on the tray will start sloshing, spilling the drink.
With the drinks on a tray that is suspended with strings you capitalize on the equivalence of inertial and gravitational mass.
Rhett demonstrates that you can run around, zigzagging all over the place: the tray will swing, but the fluid in the glass will at all times by level with respect to the tray. Since the glass is level with the tray that means the fluid will remain in the glass.
(You describe a setup with a ball on the platform, but a glass (or any beaker) with fluid is a better demonstration.)

The simplicity of the setup means that you can try this for yourself!
Do it!
Make that tray, and experience it.
Again, the reason this is effective is the equivalence of inertial mass and gravitational mass.

I think there is a limit that you need to stay below. If you accelerate too hard the tray may not rotate fast enough, and then you can topple the glass

Some more conditions:
The mass of the suspending strings must be small compared to the mass of the tray, you need the mass of the string to be negligably small.
You also ask about suspending with rigid rods, but with a hinge up top so that the tray can still swing freely
As long as the tray can swing freely then the fluid in the glass will be level with respect to the tray.
A: If the plate is fixed to the helicopter by strings, the equilibrium condition is an oscillation of an arbitrary amplitude around the vertical. You set amplitude zero as boundary conditions for t = 0.
If the helicopter accelerates to any direction horizontally, and keeps that acceleration, the new equilibrium position has now an angle $\theta$ with the vertical. But when the platform comes to that position, its angular velocity is maximum and it will continue until reaching $2\theta$. And while the acceleration of the helicopter doesn't change, the oscillations continue with an angular amplitude of $\theta$.
Supposing no friction, the ball runs back and forth with respect to the center of the platform, while  it oscillates. If it falls or not, it depends on the size of the platform, comparared to the amplitude of the oscillations. And that is a function of the acceleration.
A: I am going to assume that there is no friction between the plate and the ball and that the helicopter is accelerating horizontally with constant acceleration $a$.
If the ball does not roll along the plate then the plate must make an angle $\displaystyle \tan^{-1} \frac a g$ with the horizontal. The total force exerted by the strings on the plate must also be at an angle $\displaystyle \tan^{-1} \frac a g$ from the vertical. So the ball will stay stationary on the plate provided the total force exerted by the strings on the plate is perpendicular to the plate.
If the strings are equal in length and are attached to the helicopter at a point vertically above the centre of the plate when the plate is a rest then I think there is a symmetry argument that shows that the total force exerted by the strings will always be perpendicular to the plate, so the ball will not roll off the plate.
On the other hand, there are other arrangements where this is not the case. For example, if the strings are attached to the helicopter at points vertically above each corner of the plate then the plate will always be horizontal, and any horizontal acceleration will make the ball roll off the plate.
So the answer depends on the geometry of where the strings are attached to the helicopter.
