2.3-10: Prior to chute open, the velocity model is v'=-32-0.15v, v(0)=0.
The distance model is y'=v(t), y(0)=10000. The answers are v(20) approx -202
and y(20) approx 7084.

After chute open, the velocity model is w'=-32-1.5w, w(0)=v(20). The
distance model is x'=w(t), x(0)=y(20). Some maple solving is required,
or a graphing calculator with zoom, to find the answer t approx 326.

=====maple example=====
 x:=t->7000+120*exp(-1.5*t)-21*t;
 solve(x(t)=0,t);  # solve an equation for t

2.3-20: The model is v'=-g-r v^2, v(0)=160 where g=32 and r=1/800. This
model is solved on page 101. Do not solve the model; use formulae (13),
(14) page 101 with v0=160, g=32, rho=1/800. The symbol y(t) is the
height of the bolt measured from the ground, so y(0)=0. Ans published in
the BOB are correct.

2.3-22: The model is v'=-g+r v^2 where g=32 and r=0.075 (page 102).
Because the parachutist bails, v(0)=0 and y(0)=10000=y0, where y(t) is
the distance from the ground. Page 102 requires v(t)<0, which means
y'(t)<0 or y(t) decreasing. Opposite coordinates in which y(t) increases
and measures the distance from the plane would require a different
model, not found on page 102. Use formulae (17), (18) to answer the
questions. Do not solve the model; that is done on page 102. Formula
(17) gives 0=y0-(1/r)ln(u(t)) where y0=10000, r=0.075, u(t)=cosh(c2-t
sqrt(rg))/cosh(c2). Solve for t to find the flight time. The terminal
velocity is given by (18).

maple =======

 cosh(u) is the hyperbolic cosine, defined by cosh(u)=(exp(u)+exp(-u))/2

 sinh(u) is the hyperbolic sine, defined by sinh(u)=(exp(u)-exp(-u))/2

 tanh(u)=sinh(u)/cosh(u)

 arctanh(u) is defined by arccosh(cosh(x))=x
 Similarly for arcsinh, arccosh.

 u:=t->cosh(c2-t*sqrt(rho*g))/cosh(c2);
 eq:=10000 = (1/rho)*ln(u(t));
 solve(eq,t);  # solve an equation for t