Free fall far away from Earth (integral substitution problem)

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SUMMARY

The integral substitution problem for finding the time t in the equation v(x) = -v_1√((R/x) - (R/h)) has been successfully solved using trigonometric substitution. The solution involves substituting x with h(sin²u), which simplifies the integral significantly. The final integrand becomes (1/√R)√(hx)dx/√(h-x), allowing for straightforward integration. This method effectively resolves the initial difficulty in evaluating the integral.

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  • Understanding of integral calculus and substitution methods
  • Familiarity with trigonometric identities and substitutions
  • Knowledge of the physical context of free fall and velocity equations
  • Basic proficiency in manipulating algebraic expressions
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  • Study advanced techniques in integral calculus, focusing on trigonometric substitutions
  • Explore the applications of velocity equations in physics, particularly in free fall scenarios
  • Learn about different methods of solving integrals, including numerical integration techniques
  • Review examples of integral problems involving substitutions to reinforce understanding
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Students studying calculus, physics enthusiasts, and anyone interested in solving complex integral problems related to motion and free fall dynamics.

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[SOLVED] Free fall far away from Earth (integral substitution problem)

Homework Statement



Given:

v(x) = -v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}

Find the time t.

Homework Equations



Listed above where v_1 , R , h are all constant.

The Attempt at a Solution




v(x) = \frac{dx}{dt}

dt = \frac{dx}{\left[-v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}

t = -\frac{1}{v_1}\int\frac{dx}{\left[\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}

I'm stuck on this integral but I'm convinced it's some kind of trigonometric substitution although i can't figure out which one. I've spent hours searching the internet for help and couldn't find anything so any help would be appreciated.

Perhaps if someone could at least point out which trigonometric substitution (if any at all) i should be trying to work with i'd be extremely grateful.

Thanks in advance.
 
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R/x - R/h =R(h-x)/hx.

The integrand becomes [1/(sqrt R)]*sqrt(hx)dx/sqrt(h-x). Put x = h*(sin^2 u).

It is now solved very easily. In the num, there is only sin^2u*du.
 
Thanks a lot that seemed to solve it for me ^_^
 

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