Initial velocity required for object to have x ke at y height

[fisiks]

... factoring in fluid dynamics, not just Newtonian physics...

At a given height, the object will possesses 75% of the kinetic energy of when it initially began moving.
Mass is irrelevant.

Given variables:

-Drag coefficient
-Fluid density
-Cross-sectioned area
-Displacement
-Initial and final kinetic energy (sort of; there's not exact value, but I'm just using 1 for initial and 0.75 for final)
-Acceleration/deceleration due to gravity
-etc

Missing variables:

Time and initial velocity.

Relevant equations would be the kinetic energy equation, drag equation, and deceleration from force, as well as some basic kinematics ones.

I'm trying to create a differential equation where time is isolated on one side while velocity is removed, but I'm having trouble.

If anyone could post the differential equation, with proper steps for derivation, I'd appreciate it

 

[SteamKing]

Would you like fries with that?

I'm sorry, PF rules require you to show some effort in order to receive help. We don't take requests to provide solutions.

 

[fisiks]

Would you like fries with that?

I'm sorry, PF rules require you to show some effort in order to receive help. We don't take requests to provide solutions.

Yes, fries would be good.

Anyways, I've tried deriving it myself, but *cough* I'm embarrassed to admit it, but I got stuck deriving the kinematics part >.>

$2Vi*t+\sqrt{3}*Vi*t-\sqrt{3}*a*t^2-d=0$

 

[fisiks]

Tried to calculate what's required to reach that height first, but I don't know how to work with when a variable depends on itself.

$V=V_i-((\rho*C_d*A*V^2)/2M - G)T$

$D=\int_0 (V_i-((\rho*C_d*A*V^2)/2M - G)T)dT$

 

[Nickie]

I would like to clarify that the initial velocity required for an object to have a certain amount of kinetic energy at a certain height is not solely dependent on fluid dynamics or Newtonian physics. It is a combination of both, as well as other factors such as the object's mass, shape, and the medium it is moving through.

To derive the differential equation for this scenario, we can start with the basic equation for kinetic energy:

KE = 1/2 * mv^2

Where KE is the kinetic energy, m is the mass of the object, and v is the velocity.

Next, we need to consider the impact of fluid dynamics on the object's motion. This can be described by the drag equation:

F = 1/2 * ρ * v^2 * Cd * A

Where F is the drag force, ρ is the fluid density, Cd is the drag coefficient, and A is the cross-sectional area of the object.

We can rewrite this equation as:

F = 1/2 * ρ * v^2 * (Cd * A)

Since force is equal to mass times acceleration (F = ma), we can equate the drag force to the deceleration of the object due to fluid dynamics:

ma = 1/2 * ρ * v^2 * (Cd * A)

Solving for acceleration, we get:

a = 1/2 * ρ * (Cd * A) * v^2 / m

Now, we can use the kinematics equation:

v^2 = u^2 + 2as

Where u is the initial velocity, and s is the displacement (in this case, the height).

We can rearrange this equation to isolate the initial velocity:

u = √(v^2 - 2as)

Substituting this into the acceleration equation, we get:

a = 1/2 * ρ * (Cd * A) * [√(v^2 - 2as)]^2 / m

Simplifying, we get:

a = 1/2 * ρ * (Cd * A) * (v^2 - 2as) / m

And finally, we can rearrange this equation to isolate time (t):

t = [2m / ρ * (Cd * A)] * ∫v^2 / (v^2 - 2as) * dv