Answer
(a) $v_0 = 3.43~m/s$
(b) Please refer to the free-body diagram.
(c) $F_N = 2.2\times ~w$
Work Step by Step
(a) We can find the initial speed $v_0$.
$v_2 = v_0^2+2gy$
$v_0^2 = 0-2gy = -2gy$
$v_0 = \sqrt{-(2)(-9.80~m/s^2)(0.60~m)}$
$v_0 = 3.43~m/s$
(b) Please refer to the free-body diagram.
(c) We can find the person's acceleration during the jump. For this part of the question, we can let $v = 3.43~m/s$.
$a = \frac{v^2-v_0^2}{2y} = \frac{(3.43~m/s)^2-0}{(2)(0.50~m)}$
$a = 11.76~m/s^2$
We can find the normal force $F_N$ that the ground exerts on the person jumping.
$\sum F = ma$
$F_N - mg = ma$
$F_N = m(a+g)$
We can express $F_N$ in terms of the person's weight $w$ (which is equal to $mg$).
$F_N = \frac{m(a+g)}{mg} = \frac{a+g}{g} = \frac{(11.76~m/s^2)+(9.80~m/s^2)}{9.80~m/s^2}$
$F_N = 2.2\times ~w$
