Parachute Problem

chosen to ensure continuity of both x and v at the time of deployment:

> IC2 := { v(td) = subs( t=td, V1 ),
>  x(td) = subs( t=td, X1 ) };

IC2 := \{v( td ) = − \frac{m g}{k1} + \frac{m g e^{(− \frac{td k1}{m} )}}{k1} , x( td ) = (

−m g td k1 + m^{2} g + x0 k1^{2} − m^{2} g e^{(− \frac{td k1}{m} )} )

/k1^{2}}

When the chute is deployed, the motion is found to be

> SOL2 := dsolve( subs(k=k2,SYS) union IC2,
>  { x(t), v(t) } ):
> V2 := subs( SOL2, v(t) );
> X2 := subs( SOL2, x(t) );

V2 := − \frac{m g}{k2}

+ \frac{m g (k1 − k2 + e^{(− \frac{td k1}{m} )} k2 ) e^{(− \frac{t k2}{m} )}}{e^{(− \frac{td k2}{m} )} k2 k1}

X2 := (− m g t k2 + m^{2} g + k2 (− m^{2} g k1

+ m^{2} g k1 e^{(− \frac{td k1}{m} )} + m g k1^{2} td − k2 m g td k1 + m^{2} g k2 + k2 x0 k1^{2}

− k2 m^{2} g e^{(− \frac{td k1}{m} )}) /k1^{2} − m^{2} g

(k1 − k2 + e^{(− \frac{td k1}{m} )} k2 ) e^{(− \frac{t k2}{m} )} /(e^{(− \frac{td k2}{m} )}

k1 )) /k2^{2}

The complete solution, for all t > 0, can be written in terms of the Heaviside function as

> V := V1 + (V2-V1)*Heaviside(t-td);
> X := X1 + (X2-X1)*Heaviside(t-td):

V := − \frac{m g}{k1} + \frac{m g e^{(− \frac{t k1}{m} )}}{k1} + (− \frac{m g}{k2}

+ \frac{m g (k1 − k2 + e^{(− \frac{td k1}{m} )} k2 ) e^{(− \frac{t k2}{m} )}}{e^{(− \frac{td k2}{m} )} k2 k1}

+ \frac{m g}{k1} − \frac{m g e^{(− \frac{t k1}{m} )}}{k1}) Heaviside( t − td )

At deployment, the position and velocity are

> subs( t=60, PARAM[ns], [ X, V ] );
> evalf( " );

[\frac{5209}{4} − \frac{981}{4} e^{( −12 )}, − \frac{981}{20} + \frac{981}{20} e^{( −12 )}]

[ 1302.248493, −49.04969863 ]

In the same way, the time of impact is found to be^{2}

 > t_impact[ns]:=solve( subs(PARAM[ns],X2)=0, t);
 > 't_impact[ns]' = evalf(t_impact[ns]);

t impact_{ns} :=

\frac{5}{7} W (− \frac{e^{(42 e^{( −12 )}− \frac{331759}{981})} ( −6 + 7 e^{( −12 )} )}{e^{( −84 )}})

− 30 e^{( −12 )} + \frac{1658795}{6867}

t impact_{ns} = 241.5601768

The corresponding position and velocity are

 > subs( t=t_impact[ns], PARAM[ns], [ X, V ] ):
 > evalf( " );

[ .1 10^{−117}, −7.007142857 ]

These results show good agreement with the graphical solution; how does this motion compare with real-life? Is the free-fall terminal velocity of 49 m/s (110 mph) reasonable? Is a landing velocity of 7 m/s (15.6 mph) survivable? According to [10], terminal free-fall velocity is approximately 53.6 m/s (120 mph) and the impact velocity should be comparable to a free-fall from a five foot (1.52 m) high wall. The free-fall solution can be used to determine the “maximum survivable” impact velocity. When the parameter values are

 > PARAM[five]:=x0=152/100, m=75, k1=15, g=981/100:

the corresponding motion is

 > X5 := subs( PARAM[five], X1 );
 > V5 := subs( PARAM[five], V1 );

X5 := − \frac{981}{20} t + \frac{24677}{100} − \frac{981}{4} e^{( − 1/5 t)}

V5 := − \frac{981}{20} + \frac{981}{20} e^{( − 1/5 t)}

The duration and impact velocity of this jump are

 > T5 := fsolve( X5=0, t, 0..infinity );
 > V_max := evalf( subs( t=T5, V5 ) );

T5 := .5671998658
V max := −5.26023068

Thus, the 7 m/s landing velocity exceeds this threshold by 33%. Similar non-physical conclusions are present in each of the other problems summarized in Table 1.

The previous example illustrates the need for appropriate parameter values. But, the real problem is with the specific form of the model. The Air Force Academy Training Guide states [10, p. 22] that the “opening

\overline{3}

2Note that W is the Lambert W function[2], i.e., the branch of the function W defined by W (x)e^{W (x)} = x that is analytic at 0.