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

AI Thread Summary
The discussion focuses on calculating the initial velocity needed for an object to achieve a specific kinetic energy at a given height, factoring in fluid dynamics alongside Newtonian physics. It notes that at a certain height, the object retains 75% of its initial kinetic energy, with mass being irrelevant to the calculations. Key variables include drag coefficient, fluid density, cross-sectional area, and gravitational acceleration, while time and initial velocity are identified as missing variables. The user seeks assistance in deriving a differential equation to isolate time without including velocity, but encounters challenges with the kinematics involved. The conversation emphasizes the importance of showing effort in problem-solving within the forum's guidelines.
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... 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 :smile:
 
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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.
 
SteamKing said:
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
 
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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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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