Answer
(a) $v_0 = 4.85~m/s$
(b) $a = 16.2~m/s^2$ (directed up)
(c) $\sum F = 1470~N$
Note that $\sum F = F_N - mg$
(d) He applied an average force of 2360 N to the ground.
Work Step by Step
(a) We can find the speed $v_0$ when he left the floor.
$v_0^2 = v^2 - 2gy = 0 -2gy$
$v_0 = \sqrt{-(2)(-9.80~m/s^2)(1.2~m)}$
$v_0 = 4.85~m/s$
(b) We can use $v = 4.85~m/s$ as the final velocity for this part of the question.
$a = \frac{v-v_0}{t} = \frac{4.85~m/s-0}{0.300~s}$
$a = 16.2~m/s^2$ (directed up)
(c) We can use his weight to find his mass $m$.
$weight = mg$
$m = \frac{890~N}{9.80~m/s^2} = 90.8~kg$
$\sum F = ma = (90.8~kg)(16.2~m/s^2)$
$\sum F = 1470~N$
Note that $\sum F = F_N - mg$
(d) $\sum F = F_N - mg$
$F_N = 1470~N + 890~N = 2360~N$
He applied an average force of 2360 N to the ground.
