Max Height of Stone with Mass 0.9 kg & KE 270 J

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The problem involves a stone with a mass of 0.9 kg thrown upward with an initial kinetic energy of 270 J, facing a constant drag force of 0.9 N. To find the maximum height, the initial kinetic energy can be converted into potential energy and work done against drag. The relevant equations include KE = 1/2mv^2 and PE = mgh, with the total energy accounting for both potential energy and work done by drag. The conservation of energy principle is applied, stating that the initial kinetic energy equals the work done against drag plus the potential energy at maximum height. The solution requires calculating the initial velocity and then using energy methods or kinematic equations to determine the height.
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Homework Statement


A stone with mass m = 0.9 kg is thrown vertically upward into the air with an initial kinetic energy of 270 J. The drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.9 N. What is the maximum height reached by the stone?


Homework Equations



KE =1/2mv^2 and PE=mgh

The Attempt at a Solution



I am not sure how I should be going about this...I have KE and the mass so I figured I could plug those numbers into the KE-1/2mv^2 equation and solve for velocity, but then I am not sure how to get the height. could you please help me out? or guide me to the right equation?
 
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sheri1987 said:

Homework Statement


A stone with mass m = 0.9 kg is thrown vertically upward into the air with an initial kinetic energy of 270 J. The drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.9 N. What is the maximum height reached by the stone?


Homework Equations



KE =1/2mv^2 and PE=mgh

The Attempt at a Solution



I am not sure how I should be going about this...I have KE and the mass so I figured I could plug those numbers into the KE-1/2mv^2 equation and solve for velocity, but then I am not sure how to get the height. could you please help me out? or guide me to the right equation?
That will get you the initial velocity. Then you either have to use energy methods; or else solve for the acceleration using Newton 2, then use the kinematic motion equation to solve for the height. Are you familiar with either method?
 
The work done by the drag force is force*distance since it's constant. Add this to your other energy equations.
 
Law of conservation of energy:
Initial kinetic energy has changed to work done to overcome the drag force AND potential energy.
Considering when the stone reached the maximum height:
270 = Work done + Potential energy
 
Use Conservation of Energy... Just Like he said : (.5mv^2 = mgh + F*h) the max height is in the variable h (height) when v (velocity) = 0
 
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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