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2.3 \#22

Short Method:

$m{dv \over dt}  = -mg - {v^2 \over 1325}$

${dv \over dt} = {dv \over dx}{dx \over dt} = {dv \over dx}v


$mv{dv \over dx}  = -mg - {v^2 \over 1325}$

${mv dv \over -mg - {v^2 \over 1325}} = dx$

Let $u = -mg - {v^2 \over 1325}$, 
 $du = - {2 v dv \over 1325}$, 

$-{1325mdu \over 2u} = dx$

$-\int{1325mdu \over 2u} = \int dx$

$-{1325m \over 2} ln|u| = x + C$

$-{1325m \over 2} ln|-mg - {v^2 \over 1325}| = x + C$

$-{1325m \over 2} ln|mg + {v^2 \over 1325}| = x + C$

$x = 0, v = 20$

$-{1325m \over 2} ln|mg + {20^2 \over 1325}| = C$

$> -1325*.15/2*ln(.15*9.8 + 20^2/1325);$

                          -56.84696820



$C = -56.84696820

$v = 0, x = ?$


$-{1325m \over 2} ln|mg| - C = x$

$> -1325*.15/2*ln(.15*9.8);$

                          -38.28545108

$> -38.28545108 - (-56.84696820);$

                          18.56151712



Answer:  30 + 18.56151563 = 48.562 m
\eject

Long Method:

$ma = F = -mg - {v^2 \over 1325}$, $m = 0.15$, $g = 9.8$

$m{dv \over dt}  = -mg - {v^2 \over 1325}$

${dv \over dt}  = -g(1 +  {v^2 \over 1325mg})$

${dv \over 1 +  {v^2 \over 1325mg}}  = -g dt$

$\int{dv \over 1 +  {v^2 \over 1325mg}}  = -g \int dt$

Let $u = {v \over \sqrt{1325mg}}$, 
$du = {dv \over \sqrt{1325mg}}$,
$\sqrt{1325mg}~du = {dv}$

$\int{\sqrt{1325mg}du \over 1 +  u^2}  = -gt + C$

$\sqrt{1325mg}~tan^{-1}(u) = -gt + C$

$\sqrt{1325mg}~tan^{-1}( {v \over \sqrt{1325mg}}) = -gt + C$

$t = 0, v = 20$

$\sqrt{1325mg}~tan^{-1}( {20 \over \sqrt{1325mg}}) =  C$


$> sqrt(1325*.15*9.8);$

                          44.13332075

$> 44.1333*arctan(20/44.1333);$

                          18.77823743

C = 18.77823743


$v = 0, t = ?


$\sqrt{1325mg}~tan^{-1}( {0 \over \sqrt{1325mg}}) = -gt + C$

$0 = -gt + C$, $gt = C$, $t = {C \over g} $

$> 18.77823743/9.8;$

                          1.916146677

$t = 1.916146677$

$\sqrt{1325mg}~tan^{-1}( {v \over \sqrt{1325mg}}) = -gt + 18.77823743$

$tan^{-1}( {v \over \sqrt{1325mg}}) = {-gt + 18.77823743 \over 44.13332075}$

$ {v \over \sqrt{1325mg}} = tan({-gt + 18.77823743 \over 44.13332075}$)

$ v = 44.13332075~ tan({-gt + 18.77823743 \over 44.13332075}$)

$ {dx \over dt} = 44.13332075 ~tan({-gt + 18.77823743 \over 44.13332075})$

$ {dx} = 44.13332075 ~tan({-gt + 18.77823743 \over 44.13332075})dt$

$ {dx} = 44.13332075 
{sin(
{-gt + 18.77823743 \over 44.13332075}
)
\over cos(
{-gt + 18.77823743 \over 44.13332075}
)}
dt$

Let $u =  cos(
{-gt + 18.77823743 \over 44.13332075})$

 $du =  -{g \over 44.13332075}
sin(
{-gt + 18.77823743 \over 44.13332075})dt$

                          
$ {dx} = -{44.13332075^2 \over g}{ du \over u}$

$ \int {dx} = -\int {44.13332075^2 \over g} {du\over u}$

$x = -{44.13332075^2 \over g} ln|u| + C$

$x = -{44.13332075^2 \over g} ln|cos({-gt + 18.77823743 \over 44.13332075})| + C$

$t = 0, x = 0 (or 30)$

$> 44.1333^2/(-9.8)*ln(abs(cos(18.77823743/44.1333)));$

                          18.56151563


$C = 18.56151563$


$t= 1.91615, x = ?$

$x = -{44.13332075^2 \over 9.8} ln|cos({-9.8(1.91615) + 18.77823743 \over 44.13332075})| + 
C$

$> -44.1333^2/(9.8)*ln(abs(cos((-9.8*1.91615 +$ 
$ 18.77823743)/44.1333)));$


                              -0.



$x = 0 + 18.56151563 = 18.56151563$

Answer:  30 + 18.56151563 = 48.562 m


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