Maximum Height of Projectile Using Cons. of Energy

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SUMMARY

The maximum height of a projectile launched at a speed of 45 m/s and an angle of 56° can be calculated using the conservation of energy principle. The relevant equations include potential energy (PE) and kinetic energy (KE), specifically mgh = 1/2 mv². The vertical component of the initial velocity must be considered, which can be derived using the formula h = 0.5(v_y²)/g, where v_y is the vertical component of the initial velocity. The angle significantly influences the calculation of maximum height.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Basic knowledge of projectile motion
  • Familiarity with trigonometric functions to resolve velocity components
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Learn how to resolve initial velocity into horizontal and vertical components
  • Study the derivation of the maximum height formula for projectile motion
  • Explore the effects of different launch angles on projectile trajectories
  • Investigate the role of air resistance in real-world projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for practical examples of energy conservation in action.

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Homework Statement



A projectile is launched with a speed of 45 m/s at an angle of 56° above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.


Homework Equations


PE=KE
mgh=1/2mv^2
gh=1/2v^2
h=.5v^2/g


The Attempt at a Solution


I thought in this equation that the angle wouldn't matter, but I know I'm wrong. I thought you could just plug it the numbers to the manipulated formula h=.5v^2/g. Please help! thanks
 
Physics news on Phys.org
Yes the angle matters. You need to consider the vertical component of the initial velocity.
 

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