Projectile Motion: Solving for Distance with Integration

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SUMMARY

The discussion focuses on solving for distance in projectile motion using integration. The user correctly derives the distance formula as d = V0*√(2h/g) by integrating the horizontal velocity Vx. However, it is established that integration is unnecessary for this problem, as the SUVAT equations can be utilized to find descent time more efficiently. The key takeaway is that while integration can yield the correct answer, simpler methods exist for solving projectile motion problems.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with integration techniques
  • Knowledge of SUVAT equations
  • Basic physics concepts such as velocity and acceleration
NEXT STEPS
  • Study the SUVAT equations for solving projectile motion problems
  • Learn about the derivation and application of integration in physics
  • Explore alternative methods for calculating distance in projectile motion
  • Review examples of projectile motion problems and their solutions
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone looking to deepen their understanding of integration in motion analysis.

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Homework Statement


13.75.PNG


Homework Equations


Projectile motion.

Vy = 0
Vx = V0

The Attempt at a Solution


I integrate Vx to find d in terms of t
I found d = V0*√(2h/g)

Correct?
 
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That's the right answer, but you do not need to use integration. Just use a SUVAT formula to find the descent time.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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