Chapter 7, Problem 110P

(a)

To determine

The direction should aim to reach the banana into the monkey’s hand.

Answer to Problem 110P

The direction should aim to reach the banana into the monkey’s hand is $59.0°$.

Explanation of Solution

The vertical distance from the monkey’s hand to the person hand is $5.00\text{m}$ and the horizontal distance from the person is $3.00\text{m}$.

Write the expression to calculate the direction.

$\theta ={\tan }^{-1}\left(\frac{h}{l}\right)$

Here, $\theta$ is the direction, h is the vertical distance and l is the horizontal distance.

Substitute $5.00\text{m}$ for h and $3.00\text{m}$ for l in the above equation to calculate $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{5.00\text{m}}{3.00\text{m}}\right)\\ ={\tan }^{-1}\left(1.67\right)\\ =59.0°\end{array}$

Conclusion:

Therefore, the direction should aim to reach the banana into the monkey’s hand is $59.0°$.

(b)

To determine

The reason for why the direction is same for different values of the banana’s launching speed.

Answer to Problem 110P

The launching speed is independent of the direction since both monkey and banana would have same downward acceleration.

Explanation of Solution

As the banana throws from the person’s hand, the acceleration on the banana is downward and is equal to the acceleration due to gravity of earth. In the case of monkey, the same downward acceleration would feel on the monkey.

Thus, the downward displacement of the monkey and banana is same with a magnitude of $\frac{1}{2}g{t}^2$. Therefore, the direction would not change with respect to the launching speed of the banana from the person’s hand.

(c)

To determine

The initial speed of the banana.

Answer to Problem 110P

The initial speed of the banana is $9.99\text{m}/\text{s}$.

Explanation of Solution

The distance from the monkey to banana is $1.67\text{m}$, and acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the expression to calculate the time required to travel the distance.

$y=\frac{1}{2}g{t}^2$

Substitute $1.67\text{m}$ for y and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g in the above equation to calculate t.

$\begin{array}{c}\frac{1}{2}\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)t^2=1.67\text{m}\\ t^2=0.34\\ t=0.583\text{s}\end{array}$

Write the expression to calculate the initial speed of the banana.

$v\cos \theta =\frac{l}{t}$

Here, v is the initial speed of the banana.

Substitute $0.583\text{s}$ for t, $59.0°$ for $\theta$ and $3.00\text{m}$ for l in the above equation to calculate v.

$\begin{array}{c}v\left(\cos 59.0°\right)=\frac{3.00\text{m}}{0.583\text{s}}\\ v=\frac{3.00\text{m}}{0.583\text{s}\left(\cos 59.0°\right)}\\ =9.99\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the initial speed of the banana is $9.99\text{m}/\text{s}$.

(d)

To determine

The magnitude of distance d.

Answer to Problem 110P

The magnitude of d is $0.21\text{m}$.

Explanation of Solution

The mass of the banana is $200\text{g}$, the mass of the monkey is $3.00\text{kg}$, the vertical distance moved by the monkey is $5.33\text{m}$ and acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write expression for the conservation of momentum along horizontal direction.

$mv\cos \theta =\left(m+M\right)V$ (I)

Here, m is the mass of the banana, M is the mass of the monkey, V is the final speed of the monkey along horizontal direction.

Rewrite the above expression in terms of V.

$V=\frac{mv\cos \theta }{\left(m+M\right)}$

Substitute $200\text{g}$ for m, $3.00\text{kg}$ for M, $9.99\text{m}/\text{s}$ for v and $59.0°$ for $\theta$ in the above equation to calculate V.

$\begin{array}{c}V=\frac{\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(9.99\text{m}/\text{s}\right)\cos 59.0°}{\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)+3.00\text{kg}\right)}\\ =\frac{1.03\text{kg}\cdot \text{m}/\text{s}}{3.20\text{kg}}\\ =0.322\text{m}/\text{s}\end{array}$

Write the expression for the conservation of momentum in vertical direction.

$m\left(v\sin \theta -gt\right)+M{V}_i=\left(m+M\right)V_f$ (II)

Here, $V_i$ is the initial speed of the monkey in vertical direction and $V_f$ is the final speed of the monkey in vertical direction.

Rewrite the above equation in terms of $V_f$.

$V_f=\frac{m\left(v\sin \theta -gt\right)+M{V}_i}{\left(m+M\right)}$ (II)

Write the expression to calculate $V_i$

$V_i=-gt$

Use the above expression in (II) to rewrite.

$V_f=\frac{m\left(v\sin \theta -gt\right)+M\left(-gt\right)}{\left(m+M\right)}$

Substitute $200\text{g}$ for m, $3.00\text{kg}$ for M , $9.99\text{m}/\text{s}$ for v, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g, $0.583\text{s}$ for t and $59.0°$ for $\theta$ in the above equation to calculate $V_f$.

$\begin{array}{c}{V}_f=\frac{\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(9.99\text{m}/\text{s}\left(\sin 59.0°\right)-9.80\text{m}/{\text{s}}^{\text{2}}\left(0.583\text{s}\right)\right)+3.00\text{kg}\left(-9.80\text{m}/{\text{s}}^{\text{2}}\left(0.583\text{s}\right)\right)}{\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)+3.00\text{kg}\right)}\\ =\frac{0.57\text{kg}\cdot \text{m}/\text{s}-17.1\text{kg}\cdot \text{m}/\text{s}}{3.20\text{kg}}\\ =-5.17\text{m}/\text{s}\end{array}$

Write the expression for the time taken by the monkey to hit the ground.

$s=V_f{t}^{\prime }-\frac{1}{2}g{\left(t^{\prime }\right)}^2$

Here, $t^{\prime }$ is the time taken to reach the ground and s is the vertical distance moved by the monkey.

Substitute $-5.33\text{m}$ for s, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g and $-5.17\text{m}/\text{s}$ for $V_f$ in the above equation to calculate $t^{\prime }$.

$\begin{array}{l}-5.33\text{m}=\left(-5.17\text{m}/\text{s}\right)t^{\prime }-\frac{1}{2}\left(9.80\text{m}/{\text{s}}^{\text{2}}\right){\left(t^{\prime }\right)}^2\\ 0=4.9\text{m}/{\text{s}}^{\text{2}}{\left(t^{\prime }\right)}^2+\left(5.17\text{m}/\text{s}\right)t^{\prime }-5.33\text{m}\end{array}$

The above equation is the quadratic form. Solve the above equation using quadratic formula to calculate $t^{\prime }$.

$\begin{array}{c}0=4.9\text{m}/{\text{s}}^{\text{2}}{\left(t^{\prime }\right)}^2+\left(5.17\text{m}/\text{s}\right)t^{\prime }-5.33\text{m}\\ t^{\prime }=\frac{-5.17\text{m}/\text{s}\pm \sqrt{{\left(5.17\text{m}/\text{s}\right)}^2-4\left(4.9\text{m}/{\text{s}}^{\text{2}}\right)\left(-5.33\text{m}\right)}}{2\left(4.9\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{-5.17\text{m}/\text{s}+11.45\text{m}/\text{s}}{9.80\text{m}/{\text{s}}^{\text{2}}}\\ =0.64\text{s}\end{array}$

Write the expression to calculate the horizontal distance d.

$d=V{t}^{\prime }$

Substitute $0.64\text{s}$ for $t^{\prime }$ and $0.322\text{m}/\text{s}$ for V in the above equation to calculate d.

$\begin{array}{c}d=\left(0.64\text{s}\right)0.322\text{m}/\text{s}\\ =0.21\text{m}\end{array}$

Conclusion:

Therefore, the magnitude of d is $0.21\text{m}$.