Chapter 4, Problem 18P

(a)

To determine

The final velocity when a person jumps from 2.8 m high roof.

Answer to Problem 18P

Solution:

Final velocity of person is $V_f=7.41\text{m/s}$ .

Explanation of Solution

Given:

Initial velocity when a person jumps off the roof, $V_i=0\text{m/s}$

Acceleration during free fall, $a=g=9.8{\text{ m/s}}^{\text{2}}$

Height through which person falls, $d=2.8\text{ m}$

(b)

To determine

To find: Final velocity as person lands on the ground, $V_f=?$

Formula:

Third equation of motion:

  $V_f^2=V_i^2+2ad$

Where, $V_f$ is the final velocity, $V_i$ is the initial velocity, a is the acceleration and d is the displacement.

Calculation:

Substitute the vales and solve for final velocity.

  $\begin{array}{l}{V}_f^2=V_i^2+2ad\\ V_f^2=0^2+2\left(9.8\right)\left(2.8\right)\\ V_f=7.41\text{m/s}\end{array}$

Explanation of Solution

Conclusion: Free fall concept helps to determine the final velocity of a person as he lands on the ground is $V_f=7.41\text{m/s}$ .

c)

To determine

The average force exerted on person’s torso by his legs during deceleration.

Answer to Problem 18P

Solution:The average force exerted on person’s torso by his legs during deceleration is $2058.84\text{N}$ .

Explanation of Solution

Given:

Initial velocity when torso deceleration starts after landing on the ground, $V_i\text{'}=7.41\text{m/s}$

Final velocity when all kinds of motions ceased, $V_f\text{'}=0\text{ m/s}$

Mass of person’s leg, $m=42\text{ kg}$

(d)

To determine

To find: (I)Deceleration of person’s legs, $a\text{'}=?$

(II)Force on torso during deceleration, $F=?$

Explanation of Solution

Formula Used:

Third equation of motion:

  $V_f^2=V_i^2+2ad$

Where, $V_f$ is the final velocity, $V_i$ is the initial velocity, a is the acceleration and d is the displacement.

From Newton’s second law of motion:

  $F=ma$

Where, m is mass and a is acceleration.

Calculation:

(I)Substitute the values in the third equation of motion and solve for acceleration (a)

  $\begin{array}{l}{V}_f{\text{'}}^2=V_i{\text{'}}^2+2ad\\ 0^2={\left(7.41\right)}^2+2a(0.70)\\ a=-39.22{\text{ m/s}}^{\text{2}}\end{array}$

Thus, deceleration = $-a=39.22{\text{ m/s}}^{\text{2}}$

(II) Using Newton’s second law, calculate the average force exerted by legs on torso during deceleration.

  $\begin{array}{l}\begin{array}{l}\begin{array}{l}F=mg-ma\\ F=m(g-a) \end{array}\hfill \\ F=\left(42\right)\left(9.8-(-39.22\right)\text{N}\hfill \end{array}\\ F=2058.84\text{N}\end{array}$

Conclusion: Using Newton’s second law, the average force exerted by legs on torso during deceleration is $2058.84\text{N}$ .