A.4.3: Section 4.3 Answers
( \newcommand{\kernel}{\mathrm{null}\,}\)
1. v=−3845(1−e−5t/12);−3845ft/s
2. k=12;v=−16(1−e−2t)
3. v=25(1−e−t);25ft/s
4. v=20−27e−t/40
5. ≈17.10 ft
6. v=−40(13+3e−4t/5)13−3e−4t/5;−40ft/s
7. v=−128(1−e−t/4)
9. T=mkln(1+v0kmg);ym=y0=mk[v0−mglln(1+v0kmg)]
10. v=−64(1−e−t)1+e−t;64ft/s
11. v=αv0(1+e−βt)−α(1−e−βt)α(1+e−βt)−v0(1−e−βt);−α where α=√mgk and β=2√kgm
12. T=√mkgtan−1(v0√kmg)v=−√mgk;1−e−2√akm(t−T)1+e−2√akm(t−T)
13. s′=mg−ass+1;a0=mg
14. (a) ms′=mg−f(s)
15.
- v′=−9.8+v4/81
- vT≈−5.308m/s
16.
- v′=−32+8√|v|;vT=−16 ft/s
- From Exercise 4.3.14
- vT is the negative number such that −32+8√|vT|=0; thus, vT=−16 ft/s
17. ≈6.76miles/s
18. ≈1.47miles/s
20. α=gR2(ym+R)2