Chapter 3, Problem 103P

(a)

To determine

The maximum height made by the locust at its jump.

Answer to Problem 103P

Maximum height is $28.6\text{cm}$.

Explanation of Solution

The jump of locust is a projectile motion.

The angle of jump is $55.0°$ and the horizontal distance is $0.800\text{m}$.

At maximum height, the vertical component of velocity is zero. Write the expression for vertical component of velocity at the highest point of motion.

$v_{fy}=v_i\sin \theta -g\Delta t$

Here, the vertical component of velocity at the highest point is $v_{fy}$, magnitude of initial velocity is $v_i$, angle of projection is $\theta$, gravitational acceleration is $g$, and the time taken to reach the highest point is $\Delta t$.

Rewrite the above relation in terms of $\Delta t$ by substituting $0\text{m}/\text{s}$ for $v_{fy}$.

$\Delta t=\frac{v_i\sin \theta }{g}$ (I)

Write the equation for horizontal distance covered by locust in time $\Delta t$.

$\Delta x=\left(v_i\cos \theta \right)\Delta t$)

Here, the horizontal distance covered by locust in time $\Delta t$ is $\Delta x$.

Rewrite the above relation in terms of $\Delta t$.

$\Delta t=\frac{\Delta x}{v_i\cos \theta }$ (II)

Equate the right hand sides of equation (I) and (II).

$\frac{v_i\sin \theta }{g}=\frac{\Delta x}{v_i\cos \theta }$

Rewrite the above relation in terms of $v_i^2$.

$v_i^2=\frac{g\Delta x}{\sin \theta \cos \theta }$ (III)

Write the expression for the maximum height of a projectile.

$H=\frac{v_i^2{\sin }^2\theta }{2g}$ (IV)

Here, the maximum height is $H$.

Rewrite the above relation for $H$ by substituting the relation for $v_i^2$.

$\begin{array}{c}H={\left(\frac{g\Delta x}{\sin \theta \cos \theta }\right)}^2\frac{{\sin }^2\theta }{2g}\\ =\frac{\Delta x}{2}\tan \theta \end{array}$

Conclusion:

Substitute $0.400\text{m}$ for $\Delta x$ and $55.0°$ for $\theta$ in the above equation to find $H$.

$\begin{array}{c}H=\frac{\left(0.400\text{m}\right)}{2}\left(\tan 55.0°\right)\\ =\left(0.200\text{m}\right)\left(1.43\right)\\ =0.286\text{m}\left(\frac{{10}^{-2}\text{m}}{1\text{m}}\right)\\ =28.6\text{cm}\end{array}$

Therefore, the maximum height is $28.6\text{cm}$.

(b)

To determine

Check whether the maximum height would be smaller or larger if the locust jumps with the same initial speed at $45°$.

Answer to Problem 103P

Maximum height would be smaller than that of in $55°$.

Explanation of Solution

The jump of locust is a projectile motion.

The angle of jump is $55.0°$ and the horizontal distance is $0.800\text{m}$.

Write the final equation for $H$ calculated in part (a).

$H=\frac{\Delta x}{2}\tan \theta$

From the above equation can be seen that $H$ is directly proportional to $\tan \theta$.The value of $\tan 45°$ is lesser than that of $\tan 55°$.This means that the $H$ gained at $45°$ would be lesser that at $55°$.

Conclusion:

Therefore, the maximum height would be smaller than that of in $55°$.

(c)

To determine

Check whether the horizontal range of locust would be smaller or larger if the locust jumps with the same initial speed at $45°$.

Answer to Problem 103P

Horizontal range would be larger than that at $55°$.

Explanation of Solution

The jump of locust is a projectile motion.

The angle of jump is $55.0°$ and the horizontal distance is $0.800\text{m}$.

Write the equation for the horizontal range of a projectile.

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

Here, the horizontal rage of projectile is $R$.

From the above equation can be seen that $R$ is directly proportional to $\sin 2\theta$.

Conclusion:

Substitute $45°$ for $\theta$ to find $\sin 2\theta$.

$\begin{array}{c}\sin 2\left(45°\right)=\sin 90°\\ =1\end{array}$

Substitute $55°$ for $\theta$ to find $\sin 2\theta$.

$\begin{array}{c}\sin 2\left(55°\right)=\sin 110°\\ =0.94\end{array}$

It is found that $\sin 2\left(55°\right)$ is lesser than $\sin 2\left(45°\right)$.This means that $R$ at $45°$ would be greater than at $55°$.

Therefore, the horizontal range would be larger than that at $55°$.

(d)

To determine

The maximum horizontal rage and height achieved by locust at $45°$.

Answer to Problem 103P

Maximum horizontal range is $85.1\text{cm}$.

Maximum height is $21.3\text{cm}$.

Explanation of Solution

The jump of locust is a projectile motion.

Conclusion:

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.400\text{m}$ for $\Delta x$, and $45°$ for $\theta$ in equation (III) to find $v_i^2$.

$\begin{array}{c}{v}_i^2=\frac{\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(0.400\text{m}\right)}{\left(\sin 45°\right)\left(\cos 45°\right)}\\ =\frac{3.92{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{\left(0.707\right)\left(0.707\right)}\\ =8.34{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\end{array}$

Substitute $8.34{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}$ for $v_i^2$, $45°$ for $\theta$, and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (IV) to find $H$.

$\begin{array}{c}H=\frac{\left(8.34{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right){\left(\sin 45°\right)}^2}{2\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{4.17}{19.60}\text{m}\\ \text{=0}\text{.212}\text{m}\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)\\ =21.2\text{cm}\end{array}$

Substitute $8.34{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}$ for $v_i^2$, $45°$ for $\theta$, and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (V) to find $R$.

Therefore, the Maximum horizontal range is $85.1\text{cm}$ and the maximum height is $21.3\text{cm}$.