Chapter 4, Problem 86P

(a)

To determine

Maximum height of the locust during the its jump.

Answer to Problem 86P

Maximum height of the locust during the its jump is $\underset{\_}{0.28\text{m}}$.

Explanation of Solution

Write the expression for horizontal range of the projectile motion.

    $R=\frac{v_i^2\sin 2\theta }{g}$        (I)

Here, $v_i$ is the initial velocity, $\theta$ is the projectile angle, $g$ is the acceleration due to gravity, and $R$ is the horizontal range.

Solve equation (I) for $v_i$.

    $v_i=\sqrt{\frac{Rg}{\sin 2\theta }}$        (II)

Write the kinematic equation.

    $v_{yf}^2=v_{yi}^2+2a_{y}y$        (III)

Here, $v_{yf}$ is the final velocity in y direction, $v_{yi}$ is the initial velocity in y direction, $y$ is the vertical distance, and $a_y$ is the acceleration in y direction.

Substitute, $0$ for $v_{yf}$, $v_i\sin \theta$ for $v_{yi}$, and $H$ for $y$ in the equation (III), to find maximum height.

    $\begin{array}{c}0={\left(v_i\sin \theta \right)}^2-2gH\\ H=\frac{{\left(v_i\sin \theta \right)}^2}{2g}\end{array}$        (IV)

Conclusion:

Substitute, $0.8\text{m}$ for $R$, $55°$ for $\theta$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (II), to find $v_i$.

    $\begin{array}{c}{v}_i=\sqrt{\frac{\left(0.8\text{m}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{\sin 2\times 55°}}\\ =2.9\text{m}/\text{s}\end{array}$

Substitute, $2.9\text{m}/\text{s}$ for $v_i$, $55°$ for $\theta$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (IV), to find $H_1$.

    $\begin{array}{c}{H}_1=\frac{{\left[\left(2.9\text{m}/\text{s}\right)\left(\sin 55°\right)\right]}^2}{2\left(9.8\text{m}/{\text{s}}^2\right)}\\ =0.28\text{m}\end{array}$

Therefore, the maximum height of the locust during the its jump is $\underset{\_}{0.28\text{m}}$.

(b)

To determine

Whether the maximum height will be smaller or larger if $\theta =45°$.

Answer to Problem 86P

The maximum height will be $\underset{\_}{\text{smaller}}$ if $\theta =45°$.

Explanation of Solution

Conclusion:

Substitute, $2.9\text{m}/\text{s}$ for $v_i$, $45°$ for $\theta$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (IV), to find $H_2$.

    $\begin{array}{c}{H}_2=\frac{{\left[\left(2.9\text{m}/\text{s}\right)\left(\sin 45°\right)\right]}^2}{2\left(9.8\text{m}/{\text{s}}^2\right)}\\ =0.21\text{m}\end{array}$

Therefore, the maximum height will be $\underset{\_}{\text{smaller}}$ if $\theta =45°$.

(c)

To determine

Whether the range will be smaller or larger if $\theta =45°$.

Answer to Problem 86P

The range will be $\underset{\_}{\text{larger}}$ if $\theta =45°$ than $\theta =55°$.

Explanation of Solution

Conclusion:

Substitute, $2.9\text{m}/\text{s}$ for $v_i$, $45°$ for $\theta$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (I), to find $R_2$.

    $\begin{array}{c}{R}_2=\frac{{\left(2.9\text{m}/\text{s}\right)}^2\sin 2\left(45°\right)}{9.8\text{m}/{\text{s}}^2}\\ =0.585\end{array}$

Therefore, the range will be $\underset{\_}{\text{larger}}$ if $\theta =45°$ than $\theta =55°$.

(d)

To determine

The maximum height and range if $\theta =45°$.

Answer to Problem 86P

The maximum height if $\theta =45°$ is $\underset{\_}{0.21\text{m}}$, and maximum range is $0.858\text{m}$.

Explanation of Solution

From the part (c), it is clear that the maximum height if $\theta =45°$ is $0.21\text{m}$. And From the part (d), it is clear that the range if $\theta =45°$ is $0.858\text{m}$.

Conclusion:

Therefore, The maximum height if $\theta =45°$ is $\underset{\_}{0.21\text{m}}$, and maximum range is $\underset{\_}{0.858\text{m}}$.