a) After drawing a diagram of the situation, I realized that Felix would view the earth as a circle, at a tangent to the earth at that point, and that this circle would go out as far as the line from him, to the edge of the earth would reach (Hopefully my diagram makes sense of this). This seemed reasonable as if he were farther away, the angle between the line in the direction he falls, and the line to the edge of the earth would decrease, and thus cause a smaller viewing area.
With the knowledge of the Earth’s radius (6200km) I could use trigonometry to find the angle (labeled theta on the diagram) between Felix, and the side of the earth (assuming that Felix jumps directly above the center of mass of the earth). After finding this angle, I could then use the knowledge that Felix is 390000m above the earth, to find the scope of the earth that he could view, as he can view a circle of the earth from above. I would then divide the total area of the circle he could see, by the total area of the circle with radius 6300km which is the total viewable area of the earth to find the proportion of the earth he views.
Here are my calculations:
Opposite side to angle = 6300000m
Adjacent side to angle = 39000 + 6300000 = 6339000m
Tantheta = opposite/adjacent
Tantheta = 6300000/6339000
Tantheta = 2100/2113
Opposite side to angle = x
Adjacent side to angle = 39000
Tantheta = opposite/adjacent
2100/2113 = x/39000
x = 1748104.945m
Area = pi x r^2
= pi x 1748104.945^2
=9.6x10^12
Area = pi x r^2
= pi x 6300000^2
= 1.25 x10^14
9.6x10^12/1.25x10^14 = 0.07699
about 8% of earth’s surface.
b) For this question I could use the suvat equations.
S = 390000, u=0, a=9.8, t=?
S=ut+(at^2)/2
39000=4.9t^2
t^2=7959.2
t=89.2
= 1 minute 30s
so 7minutes and 33s earlier
c) Again, suvat equations
U=0 v=? a=0.8 t= (4x60) + 19
V= u +at
V=9.8x259
V=2538.2m/s