Answer
a. $6144\ ft$
b. $4296\ ft$
c. $B$
Work Step by Step
a. The equation for a free fall starting from zero velocity at height $h$ is $s=h-\frac{1}{2}gt^2$, where $g\approx32\ ft/s^2$ is the gravity acceleration constant, and $t$ is the time after the jump. This equation is a standard physics formula which you can get by integrating $s=\int vdt=\int -gtdt$ with an initial condition of $s(0)=h$. For skydiver-A, $t=4\ sec$ and $h=6400\ ft$, we have the altitude when his parachute open as $s(4)=6400-\frac{1}{2}(32)(4)^2=6144\ ft$
b. Similarly, for skydiver-B, $t=13\ sec$ and $h=7000\ ft$, we have the altitude when his parachute open as $s(45)=7000-\frac{1}{2}(32)(13)^2=4296\ ft$
c. The total time before landing is the time when the parachute opened plus the time for descending the remaining height with a constant speed of $16\ ft/sec$. For skydiver-A, $T_A=4+\frac{6144}{16}=388\ sec$. For skydiver-B, we also need to consider the delay of the jump; thus $T_B=45+13+\frac{4296}{16}=326.5\ sec$. Since $T_B\lt T_A$, diver-B will land first.
