I think it’s c because the other ones are just options not facts
A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

Using (2) we can rewrite this as

And using (1), we find

Substituting
into the last equation, we find the value of x:

B) 
In this case, the kinetic energy is 1/10 of the total energy:

Since we have

we can write

And so we find:

Answer:
They are 7.4m apart.
Explanation:
Here we have a parabolic motion problem. we need the time taken to land so:

considerating only the movement on Y axis:

Because we have a contant velocity motion on X axis:

and

the distance between them is given by:

Answer:
I think like 2024 or 25
Explanation:
Elon musk will probably go