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OJCO: Coer => T: lenw 239 woo 17.6: Section 2.3, Problem 23 Problem. Assume that a ball with mass 0.15199 is thrown upward with initial
velocity 20? from the roof of a building 30m high. There is a force due to air
resistance of $25122, where v is measured in %. a. Find the maximum height above the ground that the ball reaches. Solution. Although one wants an expression for the height a:(t) of the ball
as a function of time, the equation one should probably begin With is a ﬁrstorder
differential equation for the velocity v(t) of the ball. Once one solves for that,
ﬁnding the height should be simple: if; = 11(t), which will be a separable differential
equation. I l The expression for 92—: is a bit more complicated. Suppose one begins with
the convention that a positive velocity corresponds to the ball heading up. Then
the ball feels a constant acceleration due to gravity of —10;"; (it is in fact closer to g
9.8, but the exact number of the constant does not change the way the solution is
derived, it merely makes the numbers in the equation longer). Air resistance also
slows the ball, so that when it is rising, it feels a force of —ﬁv2 (and this part is
concerned only with the ball’s rise). The force acting on the ball, of course, is the product of its mass and its
acceleration (the change in velocity over time), so that mil—‘1; 2 —10m — $122. A
differential equation describing the ball’s rise is therefore: do 1 2 p d_t = ”10 ‘ (0.15)(1325) ” This is, incredibly, a separable equation, equal to: 1 dv/dt
(Toxﬁ‘ggi) — ‘1 The. left side one instantly recognizes as being a form of the derivative of
arching). » One may then integrate both sides of this equation by dt, which will show
the arctangent of some multiple of v(t) to be some multiple of —(t + C), with 0 some constant of integration. With an expression this complex — and, one must {I 2.13 P Bl remember, it will grow more complex when one looks for 50(75), the height of the ball
— it might be advantageous to use Maple’s dsolve command to do the hard work.
One might then enter:
> with (DEtools) :
'> eqnvl := diff( V(t) , t ) = 710  (1 / ( (15/100) * 1325) ) *
( v(t) ) /\ 2;
4 , ._ 2 _ _ __ 2
eqnv1.—— at” v(t): 10 795126) > dsolve( { eqnv1, v(0)= 30 }, v(t) ); v(t)——  —5\/ﬁ_tan(— \/79—5 — arctan(— —\/79—5)) 795 795 Notice that the mass of 0.15kg was entered as 15/100 — this allows Maple to
express most of the quantitieswith exact symbols, such as ——5\/7—93, rather than
‘ rough approximations such as —44.58138626. To ﬁnd the time when the ball reaches its apex, one must ﬁnd when v(t) = 0.
By looking at the expression, one may notice the ﬁrst time that v(t)—  0 will be when
tang—93¢? — arctan(795\/ﬁ—)) = O, that 1s, when 7—9—5 ——5\/79_ — arctan(7—95\/79_).
The apex, then, occurs when t = V? arctan(795 \/79—) , which is approximately when t = 1.99. To ﬁnd the height of the ball, one must solve the new initial value prob
lem 7% = v(t), 50(0) = 30. L This is again a separable differential equation; in—
deed, it is already separated — % = —5¢793tan (ﬁx/ﬁg — arctan(%5\/'E_5_)).
Integrating both sides with respect to dt then turns the equation into a:(t) =
f —5\/793tan (735m —— arctan (ﬁgﬁﬂdt. The integral of a tangent function
is the logarithm of a secant, but again this entire problem can be handled very
quickly by Maple: _ > eqnxl := diff(x(t), t)=  5 * sqrt(795) * tan( ( 2 / 795 ) *
sqrt( 795 ) * t  arctan( ( 4 / 795 ) * sqrt(795) ) ); eqnxl := gﬁt) = —5\/795 tan (—V 79  arctan(—— 795 ix/E—D 795 >.dsolve( { eqnxl, x(0) = 30 }, x(t) )i :I:(t)= gﬁ ln(\/795+4tan( 3975
WENgut) ~— Tln(11—(795\/7—95t) ))
.— —39475 111(3) — —39475 111(5) — —39475 ln (53) This isa very intimidating expression, and does not much resemble the loga—
rithm of the secant one expected to ﬁnd. If one looks carefully, though, one notices
the second logarithm term contains a 1 + tan (11)2 form, which is equal to sec(u)2.
The ﬁrst term similarly could be squared, if one uses the fact that aln (b) : 1n (ba),
providing some reassurance that if it were absolutely necessary, one could rewrite
the expression as the natural log of a secant function, plus some constant. , It is not necessary to do that for this problem, as one is interested only in the
maximum height the ball reaches. The ball will reach its maximum height when
t = 1.99, as argued above, so one need only evaluate x(t) at that time. This height,
:c(1.99) = 49.8, approximately. So the maximum height of the ball Will be just shy ' of 50 meters above the ground. b. Find the time that the ball hits the ground. Solution. Unfortunately, one cannot use exactly the same equation as in part
a — the direction of the air resistance has changed, and it now acts as a force upward
rather than downward. However, the general logic remains the same, so that the
skill one developed in the ﬁrst part makes the second easier. The ball is still accelerated downward at 10% , but now feels an accelera—
tion from air resistance that points up, at W112 . A differential equation
describing this new situation is therefore: do 2
— = —10 2
dt + 3975” Note that one knows the ball starts at the height of 48.2 meters above ground, and is at rest there. This is the starting point for a new fall, and one might sensibly
write the initial conditions asr'v(0)= 0, and so ﬁnd the position by writing‘f; — = 'v(t)
, 56(0) = 49.8 . This is a good way to set up the problem, but to exactly answer the Ma original question — at'what time does the ball hit the ground — one must remember
that theball spent 1.99 seconds climbing. 1.99 seconds must be added to the time
the ball takes to hit the ground in the below new work to correctly ﬁnd how long the ball spends in the air. Again one has a separable equation in this problem; it can be rewritten as 1 dv/dt 1 3975 2
10 2 arctcmh( mow» — ——t + C Where C is a constant of integration and arctanh is, of course, the inverse of the hyperbolic cotangent function. (Hyperbolic trigonometric functions may be
worth some review if they are completely unfamiliar. They are discussed in every
introductory calculus book. Fortunately, the derivatives and integrals of the hy
perbolic functions mirror the derivatives and integrals of the normal trigonometric functions, often without the frustration of plus or minus signs chan'ging.‘) While one can continue easily in this vein, it is quicker and certainly as accurate ' to use Maple:
> with(DEtools): q _
> eqnv2 := diff(v(t), t) = 10 + (2/3975) * v(t) /\ 2; 6 2 2
eqn’u2 .— amt) — —10 + Mow > dsolve( { eqnv2, v(Q) = 0 }, v(t) );
v(t) = —5tanh(7§gV795t)‘¢795 In order to ﬁnd when the ball hits the ground, one must ﬁnd its height, which
is the solution to the differential equation fl“? =‘ v(t) With the initial condition 22(0) = 49.8. As it was in this step Of part a, this is aseparable (indeed, already separated) ‘ ' ‘(oNTlNUEDl
’p—I 2.3Pl" equationi da: 2
ﬁt" _ —5tanh(%—5—\/795t)\/795 'One can integrate both sides with respect to t, turning the left side into ac(t)
and the right side into a log of a hyperbolic cosine. ’Or, as above, one can allow
Maple the chance to do the somewhat tedious work: > 'eqnxz := diff(x(t), t) = — 5 * tanh( (2 / 795) * sqrt(795) * ,
t) * sqrt(795); eqnx2 := $1.10 = —5tanh(%5—\/79—5t)m > dsolve( { eqnx2, x(0) = 49.8}, x(t) );, 75 2 3975 2
_x(t) = %—ln(tanh(m 79575)—1)+—4——ln(tanh(ﬁ§x/795t)+1)
3975  . . + 49.8 —— T7” There is the square root of a negative number here, which would seem to make
the number complex. This is a quirk of Maple  if one looks closely, the term inside
the ﬁrst logarithm, tanh (ﬁx/7550 — 1, is always negative: If one ventures into the
study of complexvalued functions, one can construct a deﬁnition for the natural
logarithm of a negative number, which looks like the natural log of the absolute
value of that number, plus a constant imaginary number. So While the expression
appears to be complex, in reality, it’s just a badly written real valued function. One
can write it instead, correctly, as 3975 7 2 2 ‘
x(t) = E942 log (1 —— tanh (ﬁx/7950) + ——4 log (tanh (ﬁx/79515) + 1) + 49.8 The ball will hit the ground then a:(t) = 0. Maple can quickly show that this occurs at approximately t = 3.17. ,
Therefore the ball spends 1.99 seconds rising (part a), and 3.17 seconds falling
(part b), for a total time of approximately 5.16 seconds before it hits the ground. c. Plot the graphs of velocity and position verSUS time. Compare these graphs We? with the corresponding ones in problems 21 and 22.
Solution. Plotting the velocity and position of these functions must be done
in pieces, one the rising portion and the other the fall. The velocity of the ball, plotted at speed ( § ) versus time ( s ) is: 0.2 0.4 0.6 0.! ‘1 1.2 1.4 1.6 1.8 The differences between this and the curves in problems 21 and 22 are slight.
In problem 21, one assumes that no air resistance exists. In this case; the velocity as
a linear function in time — it slopes down and is perfectly straight. Although it is not
immediately obvious from these plots, the velocity is curled a little, cupped down
during the ball’s rise, and cupped upward during its descent, reﬂecting the slowing
inﬂuence of the air. If the ball were able to fall forever without hitting ground, its 13.3351 _(
1213P'I8I _velocity would eventually ﬂatten out (to the ”terminal velocity”), whereas for the
case without air resistance its velocity would grow forever. If there were no air resistance, the ball w0uld trace out a parabola in» time, as
well. It does not do this exactly — the height is a little low throughout its trajectory,
and how low it is keeps increasing. . In problem 22, one models the problem with a linear air resistance, as opposed
to the quadratic used in this problem. Since the air resistance is linear, it has a
greater inﬂuence than this problem’s quadratic ( v2 ) air resistance when the ball
moves most slowly, near the apex of its ﬂight, but much less inﬂuence when the ball
moves rapidly. However, the ball spends little” time moving slowly, so that what one
sees is a sort of mix between the cases of no and of quadratic air resistance. The velocity, with linear air resistance, would be a curve cupped down (during
the ball’s ascent) or up (during its descent), but more slightly than the quadratic
resistance’s velocity as plotted here. (It also would approach some terminal velocity
if the ball could fall forever.) Its positiOn would alsobe a bit underneath a parabola’s, but it would be more parabolic than what is plotted here. 9'39. lﬁ ...
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 Spring '04
 Yoon
 Differential Equations, Equations

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