How Do You Calculate Ymax and Velocity at Half Ymax Using Energy Conservation?

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To calculate Ymax using energy conservation, the initial kinetic energy (Ki) is equal to the potential energy (Uf) at the maximum height, leading to the equation (1/2)mV0^2 = mgymax. Simplifying this gives Ymax = V0^2/(2g). For the velocity at half Ymax, the kinetic energy at that point is the sum of the potential energy at half the height and the remaining kinetic energy. Using the energy conservation principle, the velocity at y = 1/2 Ymax can be found by rearranging the energy equation to yield V1/2 = sqrt(V0^2 - gYmax). This approach effectively utilizes energy conservation to derive both Ymax and the velocity at half that height.
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Homework Statement


A ball is thrown vertically from y=0 with velocity V0. Using only energy considerations, find the following
a. Ymax in terms of V0.
B. The Velocity(V1/2 at y= 1/2 Ymax (in terms of V0

Homework Equations


Ki+Ui=Kf+Uf


The Attempt at a Solution

I'm not sure how to start this off to find the ymax in terms of V0
 
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The U in your problem is the potential energy mgy. The K is the kinetic (1/2)mv^2. At the initial time v=V0 and y=0. At the final time v=0 and y=ymax. Use your energy conservation to solve for ymax in terms of V0.
 
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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