Tu TN

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?

skeeter

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 ...

$\displaystyle t \approx 3.63 \, sec$

$\displaystyle v(3.63) \approx 11.82 \, m/s$