HW#3 
M&T Problem 2-33

The equation we want to solve is  y''[t] = g{(v/vterm)^2-1}   where vterm=2mg/CrhoA.   We can write this as
 	y"[t]=g{(y'[t]/vterm)^2-1} and solve for y[t]  or     v'[t]=g{(v[t]/vterm)^2-1} and solve for v[t].
Once we have v[t], we can find a[t]=y"[t] from a[t]=g{(v[t]/vterm)^2-1}.
 Initial conditions are y[0]=100m, v[0]=0,Cw=0.5, rho=1.29kg/m^3,g=9.8m/s^2

The BaseBall

\(Cw = 0.5; g = 9.8; rho = 1.29; m = 0.145; r = 0.0366; A = Pi*r^2; 

    vterm = Sqrt[2*m*g/\((Cw*rho*A)\)]\)solution1y = 
      NDSolve[{y''[t] == g*((y'[t]/vterm)^2 - 1), 
          y'[0] == 0, y[0] == 100}, y[t], {t, 0, 20}]solution1v = 
      NDSolve[{v'[t] == g*((v[t]/vterm)^2 - 1), v[0] == 0}, 
        v[t], {t, 0, 20}]

Plot[Evaluate[y[t]  /.  solution1y], {t, 0, 6}, 
      PlotRange -> {0, 100},  AxesLabel -> {"t(s)", "y(m)"}]

Just as a check, here is that analytic solution from problem 2-34 letting Beta=0.5CwRhoA.  The analytic and numerical solutions agree!

\(Cw = 0.5; g = 9.8; rho = 1.29; m = 0.145; r = 0.0366; A = Pi*r^2; \ 
    beta = 0.5*Cw*rho*A; 

    Plot[100 - m/beta*Log[Cosh[Sqrt[beta/m*g]*t]], {t, 0, 6}, \ 
      PlotRange -> {0, 100}, 
	AxesLabel -> {"t(s)", "y(m)"}]\)

Plot[Evaluate[v[t]  /.  solution1v], {t, 0, 5.2}, 
      AxesLabel -> {"t(s)", "v(m/s)"}]

Plot[
      g*((Evaluate[v[t]  /.  solution1v]/vterm)^2 - 1), {t, 0, 
        5.2}, AxesLabel -> {"t(s)", "a(m/s^2)"}]

The Ping-Pong Ball

Cw = 0.5; g = 9.8; rho = 1.29; m = 0.0024; r = 0.019; A = Pi*r^2; 
    vterm = Sqrt[2*m*g/(Cw*rho*A)]solution2y = 
      NDSolve[{y''[t] == g*((y'[t]/vterm)^2 - 1), 
          y'[0] == 0, y[0] == 100}, y[t], {t, 0, 20}]solution2v = 
      NDSolve[{v'[t] == g*((v[t]/vterm)^2 - 1), v[0] == 0}, 
        v[t], {t, 0, 20}]

Plot[Evaluate[y[t]  /.  solution2y], {t, 0, 15}, 
      PlotRange -> {0, 100},  AxesLabel -> {"t(s)", "y(m)"}]

Plot[Evaluate[v[t]  /.  solution2v], {t, 0, 15}, 
      AxesLabel -> {"t(s)", "v(m/s)"},  PlotRange -> {-10, 0}]

Plot[
      g*((Evaluate[v[t]  /.  solution2v]/vterm)^2 - 1), {t, 0, 15}, 
      AxesLabel -> {"t(s)", "a(m/s^2)"},  PlotRange -> {-10, 0}]

TheRaindrop

The mass of the raindrop is given by the density of water times the volume.  Assume 1000kg/m^3 and a sphere of radius0.003m.

Cw = 0.5; g = 9.8; rho = 1.29; r = 0.003; A = Pi*r^2; 
    m = 1000*4/3*Pi*r^3 \(		vterm = Sqrt[2*m*g/\((Cw*rho*A)\)]\)solution3y = 
      NDSolve[{y''[t] == g*((y'[t]/vterm)^2 - 1), 
          y'[0] == 0, y[0] == 100}, y[t], {t, 0, 20}]solution3v = 
      NDSolve[{v'[t] == g*((v[t]/vterm)^2 - 1), v[0] == 0}, 
        v[t], {t, 0, 20}]

Plot[Evaluate[y[t]  /.  solution3y], {t, 0, 10}, 
      PlotRange -> {0, 100},  AxesLabel -> {"t(s)", "y(m)"}]

Plot[Evaluate[v[t]  /.  solution3v], {t, 0, 10}, 
      AxesLabel -> {"t(s)", "v(m/s)"},  PlotRange -> {-15, 0}]

Plot[
      g*((Evaluate[v[t]  /.  solution3v]/vterm)^2 - 1), {t, 0, 10}, 
      AxesLabel -> {"t(s)", "a(m/s^2)"}]