## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 101: 19

#### 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.

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