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

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SUMMARY

The discussion focuses on calculating the maximum height (Ymax) and the velocity at half of that height (V1/2) for a ball thrown vertically with an initial velocity (V0) using energy conservation principles. The key equation utilized is the conservation of mechanical energy, expressed as Ki + Ui = Kf + Uf, where Ki is the initial kinetic energy, Ui is the initial potential energy, Kf is the final kinetic energy, and Uf is the final potential energy. By substituting the values for kinetic and potential energy, Ymax can be derived as Ymax = V0²/(2g), and V1/2 can be calculated as V1/2 = √(V0²/2).

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  • Understanding of basic physics concepts, specifically energy conservation.
  • Familiarity with kinetic energy (KE = 1/2 mv²) and potential energy (PE = mgy).
  • Knowledge of gravitational acceleration (g = 9.81 m/s²).
  • Ability to manipulate algebraic equations to solve for unknowns.
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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.
 

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