Proof: maximum range of an object launched off a cliff

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SUMMARY

The discussion centers on deriving the maximum range of an object launched off a cliff of height H with an initial velocity v at an angle θ from the vertical. The key formula presented is cos²(θ) = (v²)/(2v² + 2gh), which defines the optimal launch angle for achieving maximum range. The author emphasizes the importance of calculating the total time of flight, which includes the ascent to maximum height and the descent back to the original height, as well as the fall to the ground. Despite attempts to simplify the equations, the author struggles with complex quadratic expressions.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions and their applications
  • Knowledge of quadratic equations and their solutions
  • Basic concepts of gravitational acceleration (g)
NEXT STEPS
  • Study the derivation of projectile motion equations in physics
  • Learn about the effects of launch angles on range in projectile motion
  • Explore the application of quadratic equations in real-world scenarios
  • Investigate the impact of initial height on projectile trajectories
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Physics students, educators, and anyone interested in the mathematical modeling of projectile motion, particularly in scenarios involving initial height and launch angles.

pantheid
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I've been working on this for a while, with little luck. We are told that an object is on a cliff of height H, and is launched off with a velocity v at an angle of theta with the vertical. gravitational acceleration is g. To get the maximum range, it must be a certain angle theta, given by the formula:
cos^2(θ)=(v^2)/(2v^2+2gh). Prove that this is correct.



v*sin=range/t, -h=v*cos*T-.5gT^2, getting a quadtratic theorem, and deriving. No luck, I get ugly giant equations rather than that small solution.
 
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First find out how much time is needed. Do that by looking at the vertical component. The total time is the time it takes to go up to its maximum height, plus the time it takes to fall back to the original height, plus the time it takes to reach the ground.
 

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