Calculating Time of Free Fall Using Distance and Time Integrals

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Homework Statement



If you jump out of an airplane and your parachute fails to open our downward velocity t seconds after jumping is approximated for g=9.8m/sec^2 and k=.2 sec, by

v(t)=(g/k)(1-e^(-kt))


So, if you jump from 5000 meters above the ground write an equation whose solution is the number of seconds you fall before hitting the ground.



The Attempt at a Solution



This is coming from the section in my book on definite integrals but I'm not sure where to start. Thanks for any help.
 
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v(t)=x'(t) where x(t) is the vertical displacement. So to find x(t) given v(t) you would integrate v(t), right?
 
Integrating v(t) would give me the total meters traveled but if I'm looking for the number of seconds it took to fall 5000 meters I would need to integrate something else.

Or could I use the integral that gives me distance and say that t=d/v where d is the integral and v is v(t)? This was my first thought but I'm uncertain since v(t) is a function of time.
 
armolinasf said:
Integrating v(t) would give me the total meters traveled but if I'm looking for the number of seconds it took to fall 5000 meters I would need to integrate something else.

Or could I use the integral that gives me distance and say that t=d/v where d is the integral and v is v(t)? This was my first thought but I'm uncertain since v(t) is a function of time.

If you can find x(t) you can certainly try and solve x(t)=5000 for t, right? Using d/v is wrong. The speed isn't constand.
 
So then finding x(t) would just be finding the antiderivative of v(t) correct?
 
Sure.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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