if drop down an stone -that it is like a ball- into the slush and if it was soft enough stone come into the slush in some distance, if throw it away from a certain height to comes into slush and if we want stone to come into slush 2 times more than before from what hight should we throw it away?

Thought I did this yesterday

Ke at hit = Pe at top = m g h
m g h = F d
assume F from slush is constant maybe
then doubling d, the depth happens when doubling h

To determine from what height the stone should be thrown in order for it to come into the slush twice the distance compared to the previous throw, we can apply the concept of potential energy and conservation of mechanical energy.

Let's assume:

m - mass of the stone
h - initial height from which the stone is thrown
d1 - distance covered by the stone in the previous throw
d2 - distance we want the stone to cover in the current throw (twice the distance of the previous throw)
g - acceleration due to gravity (approximately 9.8 m/s^2)

When the stone is thrown from a height h, it possesses potential energy (PE) given by the equation:

PE = mgh

Where:
m = mass of the stone (kg)
g = acceleration due to gravity (m/s^2)
h = height from which the stone is thrown (m)

During the throw, the potential energy will be converted into kinetic energy (KE) as the stone falls towards the slush. Assuming no air resistance, the total mechanical energy (ME) of the system (stone) is conserved:

ME = KE + PE

The kinetic energy can be calculated using the equation:

KE = 1/2 * mv^2

Where:
v = velocity of the stone (m/s)

Since the stone is being dropped vertically, the initial velocity is zero. Therefore, the initial kinetic energy is also zero. Thus, the total mechanical energy can be simplified to:

ME = PE = mgh

For the first throw, the mechanical energy is equal to the potential energy:

ME1 = PE1 = mgh1

For the second throw, we want the stone to cover double the distance compared to the first throw. We know that the mechanical energy (ME) has to be conserved, so:

ME2 = PE2 = mgh2

Since we want the stone to cover twice the distance, we can express it as:

d2 = 2 * d1

Now, we can equate the potential energies of the first and second throws:

mgh1 = mgh2

Canceling the mass (m) on both sides, we get:

gh1 = gh2

Since the acceleration due to gravity (g) is the same in both cases, it cancels out:

h1 = h2

Therefore, to achieve twice the distance covered by the stone in the previous throw, we need to throw it from the same initial height (h) as before.

In conclusion, you don't need to change the height from which the stone is thrown if you want it to come into the slush twice the distance compared to the previous throw.