Answer
(a) The magnitude of acceleration is $4610~m/s^2$, which is equal to $470~g$.
(b) $F_N = 9.70\times 10^5~N$
$F_N = 471 \times ~weight$
(c) $t = 18.7~ms$
Work Step by Step
(a) We can convert the speed to units of m/s.
$v = (311~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 86.39~m/s$
$a = \frac{v^2-v_0^2}{2y} = \frac{0-(86.39~m/s)^2}{(2)(0.810~m)}$
$a = -4610~m/s^2$
The magnitude of acceleration is $4610~m/s^2$.
We can express the acceleration in g's.
$a = \frac{4610~m/s^2}{9.80~m/s^2} = 470~g$
(b) $\sum F = ma$
$F_N - mg = ma$
$F_N = ma+mg$
$F_N=(210~kg)(4610~m/s^2)+(210~kg)(9.80~m/s^2)$
$F_N = 9.70\times 10^5~N$
We can express this force in units of the capsule's weight.
$F_N = \frac{9.70\times 10^5~N}{(210~kg)(9.80~m/s^2)}$
$F_N = 471 \times ~weight$
(c) $t = \frac{v-v_0}{a} = \frac{0-86.39~m/s}{-4610~m/s^2}$
$t = 18.7~ms$
