What is the Speed of the Package at Maximum Height?

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SUMMARY

The discussion centers on calculating the speed of a package at its maximum height after being launched from Earth's surface at an angle of 51 degrees. The maximum height reached is equal to Earth's radius, 6380 km. The initial calculations provided by users varied significantly, with one user arriving at a speed of 11181.96 m/s, while another recalculated to 321467 km/s using gravitational potential energy equations. The consensus emphasizes the importance of applying both conservation of energy and angular momentum principles to accurately determine the launch speed and final speed at maximum height.

PREREQUISITES
  • Understanding of gravitational potential energy, specifically U = -GMe/r
  • Familiarity with kinetic energy equations, including KE = 1/2 mv^2
  • Knowledge of conservation laws, particularly conservation of energy and angular momentum
  • Basic trigonometry, especially in relation to angles and components of velocity
NEXT STEPS
  • Study the application of conservation of energy in gravitational systems
  • Learn about angular momentum and its conservation in orbital mechanics
  • Explore gravitational potential energy changes for large height variations
  • Practice problems involving projectile motion and energy conservation
USEFUL FOR

Students in physics, particularly those studying mechanics and gravitational systems, as well as educators looking for examples of energy conservation in real-world scenarios.

  • #31
It often helps to draw pictures. But yes, the angle between the radial and velocity vectors would be 39 degrees at launch. What is the angle going to be at the maximum altitude?

Think about it, sketch some pictures. It might help to bring the picture a bit closer to Earth. For example, imagine a baseball thrown from center field to home. At the top of the baseball's arc, what is the angle between the ball's velocity vector and local vertical?
 
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  • #32
Isnt it 90 degrees because at the top of the arc (or maximum altitute) there is only velocity in the horizontal direction and none in the vertical?
 
  • #33
Correct.

What's next? Can you finish this problem now?
 
  • #34
Yes that helps. I am going to rework the problem this weekend when i am studying for my test and if i run into any problems i will post here again.

Thank you for all the help.
 

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