How Do Kinetic and Potential Energies Change During a Ball's Ascent?

AI Thread Summary
A 0.50 kg ball thrown upward has an initial kinetic energy of 80 J. To determine the kinetic and potential energy when the ball has traveled three-quarters of the distance to its maximum height, calculations must be performed based on energy conservation principles. The speed of the ball at this point can be derived from its kinetic energy. At maximum height, the potential energy can be calculated by considering the initial kinetic energy and the height reached. Understanding these energy transformations is crucial for solving the problem effectively.
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Homework Statement


a .50 kg ball thrown upward has an initial Kinetic energy of 80 J.


Homework Equations



(a)what are its kinetic and potential energy when it has traveled 3 quarters of the distance to its maximum height?
(b) speed at this point?
(c) potential energy at the maximum height (assume a reference is chosen to be zero at the launch point)?

The Attempt at a Solution


pls reply...i nid it badly...thnks...im working on it now..
 
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Please show your attempt.
 
i don't have an idea honestly
 
Start by finding the maximum height. What is the kinetic energy at the maximum height?
 
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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