21E - University Physics with Modern Physics

Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm. During the jump, the person’s body from the knees up typically rises a distance of around $\mathbf{50}\mathbf{ }\mathbf{cm}$. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.(a) With what initial speed does the person leave the ground to reach a height of $\mathbf{60}\mathbf{ }\mathbf{cm}$? (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper’s weight  , what force does the ground exert on him or her during the jump?

Step-by-Step Solution

Answer

1. The initial speed is $3.429 \mathrm{m}/\mathrm{s}$.
2. The free body diagram can be expressed as,    
3. The force does ground exert on him is $2.2 \mathrm{w}$.  

Step 1: Identification of given data

The given data can be listed below.

The maximum height of an average person is,

 $\begin{array}{rcl}{\mathrm{y}}_{\mathrm{avg}}& =& 60 \mathrm{cm}\\ & =& \frac{60}{100}\mathrm{m}.\\ & =& 0.60 \mathrm{m}\end{array}$

The distance of rising a person from the knees is,

$\begin{array}{rcl}\mathrm{d}& =& 50 \mathrm{cm}\\ & =& \frac{50}{100}\mathrm{m}.\\ & =& 0.50 \mathrm{m}\end{array}$  

Step 2: Significance of equation of motion:

The equation of motion is used to calculate the motion of an object in 2-dimensional, and 3-dimensional. The equation of motion can easily calculate the position, velocity, acceleration, etc.

Step 3: (a) Determination of Initial speed.

The expression for the initial speed can be expressed as,

 

${\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{gy}}_{\mathrm{avg}}$    

 

Here, u is the initial velocity at rest, g is the acceleration due to gravity, ${\mathrm{y}}_{\mathrm{avg}}$ is the maximum height of a person.

 

Substitute 0 for u, $9.8 \mathrm{m}/{\mathrm{s}}^2$ for g, and 0.6 m for ${\mathrm{y}}_{\mathrm{avg}}$ in the above equation.

 

$\begin{array}{rcl}{\mathrm{v}}^2& =& 0+2\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.6 \mathrm{m}\right)\\ \mathrm{v}& =& \sqrt{2\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.6 \mathrm{m}\right)}\\ & =& 3.429 \mathrm{m}/\mathrm{s}\end{array}$  

 

Hence, required initial speed is $3.429 \mathrm{m}/\mathrm{s}$.

Step 4: (b) Determination of free body diagram of a person during the jump.

The free body diagram can be expressed as,

Here, F is the upward force and W is the weight acting downward when jumping.

Step 5: (c) Determination of the force.

The expression for the acceleration using equation of motion can be expressed as,

 

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{ad}$     

 

Here, a is the acceleration and d is the distance.

 

Substitute 0 for u, 0.5 m for d, $3.429 \mathrm{m}/\mathrm{s}$ for v  in the above equation.

  

$$\begin{gathered} {\left(3.429 \mathrm{m}/\mathrm{s}\right)}^2=0+2\mathrm{a}\left(0.5 \mathrm{m}\right) \\ \mathrm{a}=\frac{{\left(3.429 \mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.5 \mathrm{m}\right)} \\ =11.758 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$   

 

Hence, the acceleration is $11.758 \mathrm{m}/{\mathrm{s}}^2$. 

 

The expression for the force according to Newton’s second law in term of weight can be expressed as,

$\begin{array}{rcl}\mathrm{F}& =& \mathrm{ma}+\mathrm{w}\\ & =& \mathrm{ma}+\mathrm{mg}\\ & =& \mathrm{m}\left(\mathrm{a}+\mathrm{g}\right)\\ & =& \frac{\mathrm{w}}{\mathrm{g}}\left(\mathrm{a}+\mathrm{g}\right)\end{array}$  

 

Here w is the weight.

 

Substitute $9.8 \mathrm{m}/{\mathrm{s}}^2$ for g, $11.758 \mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{rcl}\mathrm{F}& =& \frac{\mathrm{w}}{9.8 \mathrm{m}/{\mathrm{s}}^2}\left(9.8 \mathrm{m}/{\mathrm{s}}^2+11.758 \mathrm{m}/{\mathrm{s}}^2\right)\\ & =& 2.2\mathrm{w}\end{array}$   

Hence, required force is 2.2w.