Max height of projectile using energy conservation

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SUMMARY

The maximum height of a projectile launched at 40 m/s at an angle of 60 degrees can be determined using the conservation of energy principle. The initial kinetic energy (KE) is calculated as (1/2)*m*(40 m/s)^2, while at maximum height, the kinetic energy is only due to the horizontal component (vx). The vertical component (vy) can be derived from the initial velocity, allowing the equation m*g*h = KE_initial - KE_final to be used to solve for height h.

PREREQUISITES
  • Understanding of projectile motion and energy conservation principles
  • Familiarity with kinetic energy (KE) and potential energy (PE) equations
  • Ability to decompose vectors into components (horizontal and vertical)
  • Basic knowledge of gravitational acceleration (g = 9.81 m/s²)
NEXT STEPS
  • Learn how to decompose initial velocity into horizontal and vertical components
  • Study the conservation of mechanical energy in projectile motion
  • Explore the equations of motion for projectiles
  • Practice solving problems involving maximum height and range of projectiles
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in applying energy conservation principles to real-world scenarios.

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Homework Statement



find max height of a projectile traveling at 40 m/s 60 degrees from the horizontal using conservation of energy

Homework Equations


KEi + PEi = KEf + PEf?


The Attempt at a Solution



couldnt attempt. (I know I can't be given the answer, but even just the proper equation would be handy, thanks!)
 
Last edited:
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Split the initial velocity v into x and y component, vx and vy. At launch the KE is (1/2)*m*v^2, at max height, it only has a vx component, so KE=(1/2)*m*(vx)^2. Set the KE difference equal to the PE difference m*g*h and solve for height h.
 

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