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 May 31st, 2015, 05:19 PM #1 Newbie   Joined: May 2015 From: Waterloo, Canada Posts: 2 Thanks: 0 Math Focus: Calculus A math problem that involves first order ODE Hi all, I came across this problem today. This is my attempted solution. As you can see, I stumbled upon an equation that is in implicit form. I found C by substituting in the initial values given, however unable to find v when y = 0 because it is in implicit form. Am I doing this wrong?
 June 2nd, 2015, 01:09 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,886 Thanks: 1505 let the downward direction be positive with $y_0 = 0$, $v_0=0$, and $y_f=30$ $\displaystyle m \frac{dv}{dt} = mg - kv$ $\displaystyle \frac{dv}{dt} = g - \frac{k}{m}v$ $\displaystyle \frac{dv}{dt} = g - 0.8v$ $\displaystyle \frac{dv}{g-0.8v} = dt$ $\displaystyle \ln|g-0.8v| = -0.8t + C$ $\displaystyle g-0.8v = Ae^{-0.8t}$ $\displaystyle v(0) = 0 \implies A = g$ ... $\displaystyle 0.8v = g(1 - e^{-0.8t})$ $\displaystyle v = \frac{5g}{4}(1 - e^{-0.8t})$ ... note terminal velocity, $\displaystyle v_T = \frac{5g}{4} \, m/s$ $\displaystyle \frac{dy}{dt} = \frac{5g}{4}(1 - e^{-0.8t})$ $\displaystyle y = \frac{5g}{4}(t + 1.25e^{-0.8t} + C)$ $\displaystyle y(0) = 0$ $\displaystyle 0 = \frac{5g}{4}(1.25 + C) \implies C = -1.25$ $\displaystyle y = \frac{5g}{4}[t + 1.25(e^{-0.8t} - 1)]$ $\displaystyle 30 = \frac{5g}{4}[t + 1.25(e^{-0.8t} - 1)]$ using $\displaystyle g = 9.8 \, m/s^2$ and solving with a calculator ... $\displaystyle t \approx 3.63 \, sec$ $\displaystyle v(3.63) \approx 11.82 \, m/s$ Thanks from Tu TN

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