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30, c = 30 and x(t) = 20t - (g/2)t + 30. At the maximum
height v(tm) = 0 and thus
tm = 20/9.8 = 2.04 sec., which when substituted in the equation for x(t) yields the maximum height.
21b. At the ground x(tg) = 0 and thus 20tg - 4.9tg + 30 = 0.
22. The I.V.P. in this case is m = -----------------v - mg, v(0) = 20,
where the positive direction is measured upward. dv
2 4a. The I.V.P. is m = mg - .75v, v(0) = 0 and v is
measured positively downward. Since m = 180/32, the D.E.
dv 2 -2t/15
becomes = 32 - -----------v and thus v(t) = 240(1-e ' ) so
that v(10) = 176.7 ft/sec.
24b. Integration of v(t) as found in (a) yields
x(t) = 240t + 1800(e 2t/15-1) where x is measured positively down from the altitude of 5000 feet. Set
2 . 3
t = 10 to find the distance traveled when the parachute opens.
24c. After the parachute opens the I.V.P. is m = mg-12v,
v(0) = 176.7, which has the solution
v(t) = 161.7e + 15 and where t = 0 now represents
the time the parachute opens. Letting t^<^> yields the limiting velocity.
24d. Integrate v(t) as found in (c) to find
x(t) = 15t - 75.8e-32t/15 + C2. C2 = 75.8 since x(0) = 0,
x now being measured from the point where the parachute opens. Setting x = 3925.5 will then yield the length of time the skydiver is in the air after the parachute opens.
26a. Again, if x is measured positively upward, then Eq.(4) of
Sect.1.1 becomes m = -mg - kv.
mg mg -kt/m
26b.From part (a) v(t) = - + [v0 + --]e / . As k ^ 0
k 0 k
this has the indeterminant form of - + . Thus rewrite
v(t) as v(t) = [-mg + (v0k + mg)e-kt/m ]/k which has the
indeterminant form of 0/0, as k ^ 0 and hence L'Hopital's Rule may be applied with k as the variable.
27a. The equation of motion is m(dv/dt) = w-R-B which, in this problem, is
4 3 4 3 4 3
pa p(dv/dt) = pa pg - 6pmav - Pg. The limiting
velocity occurs when dv/dt = 0.
27b. Since the droplet is motionless, v = dv/dt = 0, we have
4 3 4 3
the equation of motion 0 = ()pa pg - Ee - ()pa pg,
where p is the density of the oil and p' is the density
of air. Solving for e yields the answer.
28. All three parts can be answered from one solution if k represents the resistance and if the method of solution of Example 4 is used. Thus we have dv dv
m = mv----- = mg - kv, v(0) = 0, where we have assumed
the velocity is a function of x. The solution of this
I.V.P. involves a logarithmic term, and thus the answers to parts (a) and (c) must be found using a numerical procedure.
22 29b. Note that 32 ft/sec = 78,545 m/hr .
30. This problem is the same as Example 4 through Eq.(29).
In this case the I.C. is v(^R) = vo, so c =
2 1+X '
The escape velocity,ve, is found by noting that vo >
in order for v to always be positive. From Example 4, the escape velocity for a surface launch is ve(0) = V"2gR . We want the escape velocity of xo = XR to have the relation ve(^R) = .85ve(0), which yields
X = (0.85)-2 - 1 @ 0.384. If R = 4000 miles then xo = XR = 1536 miles.
31b. From part a) = v = ucosA and hence dt
x(t) = (ucosA)t + d1. Since x(0) = 0, we have d1 = 0 and
x(t) = (ucosA)t. Likewise ---------- = -gt + usinA and
therefore y(t) = -gt /2 + (usinA)t + d2. Since y(0) = h we have d2 = h and y(t) = -gt /2 + (usinA)t + h.
31d. Let t,. be the time the ball reaches the wall. Then
x(tw) = L = (ucosA)tw and thus tw = --------------. For the ball
w w w ucosA
to clear the wall y(tw) > H and thus (setting L
tw = -------, g = 32 and h = 3 in y) we get