#### Answer

(a) The velocity just before striking the ground is 7.4 m/s.
(b) The average force exerted on the torso by the legs is 2100 N directed upward.

#### Work Step by Step

(a) In the air:
$v^2 = v_0^2 + 2ay$
$v = \sqrt{2gy}$
$v = \sqrt{(2)(9.8 ~m/s^2)(2.8 ~m)}$
$v = 7.4 ~m/s$
The velocity just before striking the ground is 7.4 m/s.
(b) On the ground:
$a = \frac{v^2 - v_0^2}{2y}$
$a = \frac{0-(7.4 ~m/s)^2}{(2)(0.70 ~m)}$
$a = -39.1 ~m/s^2$
The magnitude of deceleration is $39.1 ~m/s^2$
$\sum F = ma$
$F_N - mg = ma$
$F_N = ma+mg$
$F_N = (42 ~kg)(39.1 ~m/s^2)+(42~kg)(9.80~m/s^2)$
$F_N = 2100~N$
The average force exerted on the torso by the legs is 2100 N directed upward.