Answer:
(a) 8.4375 s
(b) 1409.06 ft
(c) 40.55 seconds
Explanation:
v’(t)=a(t)=60t therefore
[tex]v(t)=\int 60t=60\frac {t^{2}}{2}=30t^{2}[/tex]
[tex]s(t)=\int v=\int 30t^{2}=30\frac {t^{3}}{3}=10t^{3}[/tex]
For first three seconds
[tex]S(3)=10*3^{3}=270[/tex]
Since motion is vertical and we take upward direction as positive, acceleration is negative hence
a(t)=-32 and [tex]v(t)=\int -32=-32t+c[/tex] but since at t=0 then v=270 hence c=270
-32.2t+270=0 hence
t=270/32=8.4375 s
(b)
[tex]s(t)=-16t^{2}+270t+B[/tex] and since s(0)=270 then
[tex]s(t)=-16t^{2}+270t+270[/tex]
[tex]s(8.4375)=-16(8.4375)^{2}+270(8.4375)+270\approx 1409.06 ft[/tex]
(c)
v(15)=-32(14)+270=-178 ft/s
[tex]s(14)=-16(14)^{2}+270(14)+270=914 ft[/tex]
Average speed in the next 5 seconds becomes [tex]\frac {-178-18}{2}=-98 ft/s and s(5)=98(5)=490 ft[/tex]
The altitude is 914-490=424 ft
[tex]Time=\frac {424}{18}=23.6 seconds[/tex]
Time when rocket hits ground
Total time=3+14+23.6=40.55 seconds
