Chapter 9, Problem 20P

(a)

To determine

The impulse athlete receives from the platform.

Answer to Problem 20P

The impulse athlete receives from the platform is $981\text{N}\cdot \text{s}$ .

Explanation of Solution

Impulse experienced by the athlete is caculated from expression of impulse force law .

Write the expression for the impulse force law.

    $I={\displaystyle \underset{t_i}{\overset{t_f}{\int }}Fdt}$                                                                                                       (I)

Here , $I$ is the impulse experienced by the athlete, $F$ is the force, $t$ is time interval, $t_i$ is the initial time and $t_f$ is the final time interval.

Conclusion:

Substitute $9200t-11500t^2$ for $F$ , $0\text{s}$ for $t_i$ and $0.800\text{s}$ for $t_f$ in  equation (I).

    $\begin{array}{c}I={\displaystyle \underset{0}{\overset{0.8}{\int }}\left(9200t-11500t^2\right)dt}\\ ={\left[\frac{9200t^2}{2}-\frac{11500t^3}{3}\right]}_0^{0.8}\\ =4600\left[{\left(0.8\right)}^2-{\left(0\right)}^2\right]-\frac{11500}{3}\left[{\left(0.8\right)}^3-{\left(0\right)}^3\right]\\ =981\text{N}\cdot \text{s}\end{array}$

Thus, the impulse athlete receives from the platform is $981\text{N}\cdot \text{s}$.

(b)

To determine

The velocity with which it reaches the platform.

Answer to Problem 20P

The velocity with which it reaches the platform is $3.43\text{m/s, down}$.

Explanation of Solution

The athlete jumps on a plateform this converts potential energy to kinetic energy.

Write the expression for the conservation of energy.

    $K_i+U_i=K_f+U_f$                                                                                         (I)

Here $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy and $U_f$ is the final potential energy.

Write the expression for the initial kinetic energy.

    $K_i=\frac{1}{2}m{v}_i^2$

Here, $m$ is the mass of the athlete, $v_i$ is the initial velocity of athlete.

Initially athlete jumps from the rest so initial velocity is zero.

Substitute $0\text{m/s}$ for $v_i$ in above equation.

    $K_i=0\text{J}$                                                                                                         (II)

Write the expression for the initial potential energy.

    $U_i=mg{h}_i$                                                                                                   (III)

Here, $g$ is the acceleration due to gravity and $h_i$ is the initial height.

Write the expression for the final kinetic energy.

    $K_f=\frac{1}{2}m{v}_f^2$                                                                                                (IV)

Here, $v_f$ is the initial velocity of athlete.

Write the expression for the final potential energy.

    $U_f=mg{h}_f$                                                                                                   (V)

Here, $h_f$ is the final height.

Substitute $0\text{J}$ for $K_i$ , $mg{h}_i$ for $U_i$, $\frac{1}{2}m{v}_f^2$ for $K_f$ and $mg{h}_f$ for $U_f$.

    $0+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f$

Substitute $0\text{m}$ for $h_f$ in above equation and solve for $v_f$.

    $v_f=\sqrt{2g{h}_i}$                                                                                                  (VI)

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $0.600\text{m}$ for $h_i$ in equation (VI).

    $\begin{array}{c}{v}_f=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(0.600\text{m}\right)}\\ =3.43\text{m/s, down}\end{array}$

Thus, the velocity with which it reaches the platform is $3.43\text{m/s, down}$.

(c)

To determine

The velocity with which athlete leaves the platform.

Answer to Problem 20P

The velocity with which athlete leaves the platform is $3.83\text{m/s,}\text{up}$

Explanation of Solution

The athlete falls on the platform as a result impulse is produced.

Write the expression for the net impulse.

    $I_{net}=I_g+I_p$                                                                                              (VII)

Here, $I_{net}$ is the net impulse, $I_g$ is the impulse due the gravity and $I_p$ is the impulse on the plateform.

Write the expression for the impulse due to gravity.

    $I_g=\left(-mg\right)\Delta t$                                                                                           (VIII)

Here, $\Delta t$ is the time duration.

Write the expression for the impulse momentum law.

    $\Delta p=I_{net}$                                                                                                      (IX)

Here, $\Delta p$ is the change in momentum.

Write the expression for the change in momentum.

    $\Delta p=p_f-p_i$                                                                                                (X)

Here, $p_i$ is the initial momentum and $p_f$ is the final momentum.

Write the expression for the initial momentum.

    $p_i=m{v}_i$

Write the expression for the final momentum.

    $p_f=m{v}_f$

Substitute $m{v}_f$ for $p_f$ and  $m{v}_i$ for $p_i$ in the equation (X).

    $\Delta p=m\left(v_f-v_i\right)$

Substitute $m\left(v_f-v_i\right)$ for $\Delta p$ in the equation (IX).

    $I_{net}=m\left(v_f-v_i\right)$

Substitute $m\left(v_f-v_i\right)$ for $I_{net}$  and $\left(-mg\right)\Delta t$ for $I_g$ in equation (VII).

    $m\left(v_f-v_i\right)=\left(-mg\right)\Delta t+I_p$                                                                        (XI)

Simplify the above equation for $v_f$.

  $v_f=\frac{\left(-mg\right)\Delta t+I_p}{m}+v_i$                                                                              (XI)

Conclusion:

Substitute $65\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $0.800\text{s}$ for $\Delta t$ , $981\text{N}\cdot \text{s}$ and $\left(-3.43\text{m/s}\right)$ for $v_i$ in the equation (XI).

    $\begin{array}{c}{v}_f=\frac{\left(-\left(65\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\right)\left(0.800\text{s}\right)+\left(981\text{N}\cdot \text{s}\right)}{\left(65\text{kg}\right)}+\left(-3.43\text{m/s}\right)\\ =3.83\text{m/s,}\text{up}\end{array}$

Thus, the velocity with which athlete leaves the platform is $3.83\text{m/s,}\text{up}$.

(d)

To determine

The height athlete jumps from the platform.

Answer to Problem 20P

The height athletes jumps from platform is $0.748\text{m}$.

Explanation of Solution

The athlete jumps from the plateform to some height. Hence, kinetic energy is converted to gain potential energy.

Write the expression for the conservation of energy.

    $K_p+U_p=K_h+U_h$                                                                                    (XII)

Here, $K_p$ is the initial kinetic energy at platform , $U_p$ is the initial potential energy at platform , $K_h$ is the final kinetic energy at maximum height and $U_h$ is the final potential  energy at maximum height.

Write the expression for the initial kinetic energy at plateform .

    $K_p=\frac{1}{2}m{v}_p^2$

Here, $v_p$ is the initial velocity of athlete at platform.

Initially athlete jumps from the rest so initial velocity is zero.

Write the expression for the initial potential energy on platform.

    $U_p=mg{h}_p$

Here, $h_p$ is the initial height at plateform.

Initial height at plateform is zero.

Substitute $0\text{m}$ for $h_p$ in above equation.

    $U_p=0$

Write the expression for the final kinetic energy at max height.

    $K_h=\frac{1}{2}m{v}_h^2$

Here, $v_h$ is the final velocity of athlete at max height.

Substitute $0\text{m/s}$ for $v_h$ in above equation.

  $K_h=0\text{J}$

Write the expression for the final potential energy at max height.

    $U_h=mg{h}_{\max }$

Here $h_{\max }$ is the maximum height athlete jump from the plateform.

Substitute $0\text{J}$ for $U_p$  , $\frac{1}{2}m{v}_p^2$ for $K_p$ , $0\text{J}$ for $K_h$ and $mg{h}_{\max }$ for $U_h$ in equation (XII).

    $0\text{ J}+\frac{1}{2}m{v}_p^2=0\text{ J}+mg{h}_{\max }$

Simplify the above expression for value of $h_{\max }$.

    $h_{\max }=\frac{v_p^2}{2g}$                                                                                                 (XIII)

Conclusion:

Substitute $3.83\text{m/s}$ for $v_p$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in equation (XIV).

    $\begin{array}{c}{h}_{\max }=\frac{{\left(3.83\text{m/s}\right)}^2}{2\left(9.8\text{m/s}\right)}\\ =0.748\text{m}\end{array}$

Thus, the height athletes jumps from platform is $0.748\text{m}$.