Throw a ball up how high does it go

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An object projected upward at 2.98 km/s will reach a maximum height where gravitational potential energy and kinetic energy balance. The initial kinetic energy can be expressed as (1/2 mv^2), while the potential energy at height h is approximated as mgh for small heights. However, at this speed, the change in gravitational force becomes significant, requiring the use of the formula for gravitational potential energy, -GmM/r. The discussion emphasizes that the height reached will affect the gravitational pull experienced by the object. Understanding these energy transformations is crucial for accurately calculating the maximum height.
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An object is projected straight upward from the surface of Earth with an initial speed of 2.98 km/s. What is the maximum height it reaches?


Ok i know this seams like a really easy problem but i can't figure it out... since this question is from our chapter on gravity i have trued usuing is ke initial = potential energy final...
ke initial being(1/2 m v^2) and potential energy final being: (GMm/r^2)^1/2 M is the mass of the Earth and m is the mass of the ball
 
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You're not going to throw the ball so high that its distance to the center of mass of the Earth will change by a significant percentage. g is nearly constant. Therefore PE at height h will be mgh.
 
mikelepore said:
You're not going to throw the ball so high that its distance to the center of mass of the Earth will change by a significant percentage. g is nearly constant. Therefore PE at height h will be mgh.

he said 2.98 km/s. that will get you high enough for g get significantly smaller.

the potential energy for a mass at distance r from the Earth is -GmM/r, if the potential at infinity is 0, so there is both an initial and a final potential energy.
 
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