Chapter 3, Problem 64P

(a)

To determine

At what horizontal distance from seed pod does the seed hit the ground?

Answer to Problem 64P

The horizontal distance from seed pod does the seed hit the ground is $0.57\text{m}$.

Explanation of Solution

Write the equation to find the displacement of the seed.

$\Delta y=v_{iy}\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$                                                                                     (I)

Here, $\Delta y$ is the displacement, $v_{iy}$ is the initial velocity, $\Delta t$ is the change in time, $a_y$ is the acceleration

Rewrite equation (I) to find $\Delta t$

$\Delta t=\sqrt{\left(\frac{\Delta y}{a_y}\right)}$ (II)

Write the equation to find the horizontal displacement.

$\Delta x=v_x\Delta t$ (III)

Here, $\Delta x$ is the horizontal displacement, $v_x$ is the velocity, $\Delta t$ is the change in time

Conclusion

Substitute $-1.1\text{m}$ for $\Delta y$ , $0\text{m}/\text{s}$ for $v_{iy}$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (I)

$-1.1\text{m=0+0}\text{.5}\left(-9.8\text{m}/{\text{s}}^{\text{2}}\right){\left(\Delta t\right)}^2$

Solve above equation to get $\Delta t$

$\begin{array}{c}\Delta t=\sqrt{\frac{2.2{\text{ms}}^2}{9.8\text{m}/{\text{s}}^{\text{2}}}}\\ =0.474\text{s}\end{array}$

Substitute $0.474\text{s}$ for $\Delta t$ and $1.2\text{m}/\text{s}$ for $v_x$ in equation (III) to get $\Delta x$

$\begin{array}{c}\Delta x=\left(1.2\text{m}/\text{s}\right)\left(0.474\text{s}\right)\\ =0.57\text{m}\end{array}$

Therefore, The horizontal distance from seed pod does the seed hit the ground is $0.57\text{m}$.

(b)

To determine

At what horizontal distance from seed pod does the seed hit the ground if the seed is launched at $17°$ from horizontal?

Answer to Problem 64P

The horizontal distance from seed pod does the seed hit the ground is $0.59\text{m}$.

Explanation of Solution

Write the equation to find the final velocity of seed.

$v_{fy}=\sqrt{v_{yi}{}^2+2a_y\Delta t}$ (IV)

Here, $v_{fy}$ is the final velocity, $v_{iy}$ is the initial velocity, $a_y$ is the acceleration, $\Delta t$ is the time

Write the equation to find the difference in final and initial velocities.

$v_{yf}-v_{iy}=a_y\Delta t$ (V)

Here, $v_{fy}$ is the final velocity, $v_{iy}$ is the initial velocity, $a_y$ is the acceleration, $\Delta t$ is the time

Rewrite equation (II) to get $\Delta t$

$\Delta t=\frac{v_{yf}-v_i\sin {\theta }_i}{-g}$ (VI)

Write the equation to find the horizontal displacement.

$\Delta x=v_x\Delta t$ (III)

Here, $\Delta x$ is the horizontal displacement, $v_x$ is the velocity, $\Delta t$ is the change in time

Conclusion:

Substitute $1.2\text{m}/\text{s}$ for $v_{iy}$ , $17°$ for $\theta$ , $-9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $-1.1\text{m}$ for $\Delta y$ in equation (I) to get $v_{fy}$

$\begin{array}{c}{v}_{fy}=\sqrt{{\left(1.2\text{m}/\text{s}\right)}^2{\sin }^2\left(17°\right)+2\left(-9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-1.1\text{m}\right)}\\ =-4.68\text{m}/\text{s}\end{array}$

Substitute $-4.68\text{m}/\text{s}$ for $v_{fy}$ , $1.2\text{m}/\text{s}$ for $v_i$ , $17°$ for $\theta$, $-9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (III) to get $v_{fy}$

$\begin{array}{c}\Delta t=\frac{-4.68\text{m}/\text{s-}\left(1.2\text{m}/\text{s}\right)\sin 17°}{-9.8\text{m}/{\text{s}}^{\text{2}}}\\ =0.439\text{s}\end{array}$

Substitute $1.2\text{m}/\text{s}$ for $v_x$ and $0.513\text{s}$ in equation (IV) to get $\Delta x$

$\begin{array}{c}\Delta x=\left(1.2\text{m}/\text{s}\right)\cos 17°\left(0.513\text{s}\right)\\ =0.59\text{m}\end{array}$

Therefore, The horizontal distance from seed pod does the seed hit the ground is $0.59\text{m}$.

(c)

To determine

Whether the air resistance is negligible if the horizontal distance measured is $0.44\text{m}$ ?

Answer to Problem 64P

The air resistance is not negligible.

Explanation of Solution

The measured distance of $0.44\text{m}$ is less than the $0.59\text{m}$. That is the range or horizontal distance travelled by seed pod is reduced. Hence there is a resistive force present. Therefore, air resistance is not negligible.