Chapter 4, Problem 46P

(a)

To determine

The horizontal distance from seed pod to point where it hits the ground.

Answer to Problem 46P

The horizontal distance is $0.56\text{m}$.

Explanation of Solution

Write the equation to find the horizontal distance from seed pod to point where it hits the ground.

    $\Delta x=v_{\text{ix}}{t}_{\text{f}}$

Here, $\Delta x$ is the horizontal distance, $v_{\text{ix}}$ is the horizontal component of initial velocity and $t_{\text{f}}$ is the time of flight.

Since the seed is launched horizontally, initial velocity is equal to the horizontal component of velocity itself. Write the expression for $v_{\text{ix}}$.

    $v_{\text{ix}}=v$

Here, $v$ is the initial velocity and $\theta$ is the angle of projection.

Rewrite the expression for $\Delta x$ by substituting the above relation.

    $\Delta x=v{t}_{\text{f}}$

Write the equation to find $t_{\text{f}}$ from vertical component of displacement.

  $\Delta y=v_{\text{iy}}{t}_{\text{f}}+\frac{1}{2}{a}_y{t}_{\text{f}}^{\text{2}}$

Here, $\Delta y$ is the vertical displacement, $v_{\text{iy}}$ is the vertical component of initial velocity and $a_y$ is the y-component of acceleration.

Since the seed is launched horizontally, vertical component of initial velocity is zero. y-component of acceleration is gravitational acceleration only.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $v_{\text{iy}}$, $1.1\text{m}$ for $\Delta y$ and $-9.8\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in the equation to find $t_{\text{f}}$.

  $\begin{array}{c}1.1\text{m}=\left(0\text{m}/\text{s}\right)t_{\text{f}}+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)t_{\text{f}}^{\text{2}}\\ =\left(4.9\text{m}/{\text{s}}^{\text{2}}\right)t_{\text{f}}^{\text{2}}\\ t_{\text{f}}=\sqrt{\frac{1.1\text{m}}{4.9\text{m}/{\text{s}}^{\text{2}}}}\\ =0.47\text{s}\end{array}$

Substitute $1.2\text{m}/\text{s}$ for $v$ and $0.47\text{s}$ for $t_{\text{f}}$ in the above equation to find $\Delta x$.

    $\begin{array}{c}\Delta x=\left(1.2\text{m}/\text{s}\right)\left(0.47\text{s}\right)\\ =0.56\text{m}\end{array}$

Thus, the horizontal distance is $0.56\text{m}$.

(b)

To determine

The horizontal distance from seed pod to point where it hits the ground if it is projected at $17°$ from the horizontal.

Answer to Problem 46P

The horizontal distance is $0.59\text{m}$.

Explanation of Solution

Write the expression for $v_{\text{ix}}$.

    $v_{\text{ix}}=v\cos \theta$

Here, $\theta$ is the angle of projection.

Rewrite the expression for $\Delta x$ by substituting the above relation.

    $\Delta x=\left(v\cos \theta \right)t_{\text{f}}$        (I)

$t_{\text{f}}$ is divided into parts. First one is the time taken by seed reach the maximum height. Second part is the time take to travel from maximum height to ground.

Consider the first part. Write the equation to find the maximum height of a projectile.

    $h=\frac{v^2{\sin }^2\theta }{2g}$        (II)

Write the equation to find the time taken to reach the height point of motion.

    $T=\frac{v\sin \theta }{g}$        (III)

Here, $T$ is the time take to travel from maximum height to ground.

At the maximum height of projectile motion, vertical component of velocity is zero. Find the time taken to reach the ground from maximum height.

    $\Delta y=v_{\text{iy}}{t}_{\text{f}}+\frac{1}{2}{a}_y{t}_{\text{down}}^{\text{2}}$        (IV)

Here, $t_{\text{down}}$ is the time taken to reach the ground from maximum height.

Write the equation for $t_{\text{f}}$.

    $t_{\text{f}}=T+t_{\text{down}}$

Conclusion:

Substitute $1.2\text{m}/\text{s}$ for $v$, $17°$ for $\theta$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (II) to find $h$.

    $\begin{array}{c}h=\frac{{\left(1.2\text{m}/\text{s}\right)}^2{\left(\sin 17°\right)}^2}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ =6.3\times {10}^{-3}\text{m}\end{array}$

Substitute $1.2\text{m}/\text{s}$ for $v$, $17°$ for $\theta$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (III) to find $h$.

  $\begin{array}{c}T=\frac{\left(1.2\text{m}/\text{s}\right)\left(\sin 17°\right)}{9.8\text{m}/{\text{s}}^{\text{2}}}\\ =0.04\text{s}\end{array}$

Substitute $1.1\text{m+}6.3\times {10}^{-3}\text{m}$ for $\Delta y$, $0\text{m}/\text{s}$ for $v_{\text{iy}}$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (IV) to find $t_{\text{down}}$.

    $\begin{array}{c}1.1\text{m+}6.3\times {10}^{-3}\text{m}=\left(0\text{m}/\text{s}\right)t_{\text{f}}+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)t_{\text{down}}^{\text{2}}\\ 1.1063\text{m}=\left(4.9\text{m}/{\text{s}}^{\text{2}}\right)t_{\text{down}}^{\text{2}}\\ t_{\text{down}}=\sqrt{\frac{1.1063\text{m}}{4.9\text{m}/{\text{s}}^{\text{2}}}}\\ =0.475\text{s}\end{array}$

Substitute $0.475\text{s}$ for $t_{\text{down}}$ and $0.04\text{s}$ for $T$ in the equation to find $t_{\text{f}}$.

    $\begin{array}{c}{t}_{\text{f}}=0.04\text{s}+0.475\text{s}\\ \text{=0}\text{.515}\text{s}\end{array}$

Substitute $1.2\text{m}/\text{s}$ for $v$, $17°$ for $\theta$ and $\text{0}\text{.515}\text{s}$ for $t_{\text{f}}$ in equation (I) to find $\Delta x$.

    $\begin{array}{c}\Delta x=\left(1.2\text{m}/\text{s}\right)\left(\cos 17°\right)\left(\text{0}\text{.515}\text{s}\right)\\ =0.59\text{m}\end{array}$

Thus, the horizontal distance is $0.59\text{m}$.

(c)

To determine

Check whether the air resistance is negligible in second case or not.

Answer to Problem 46P

Role of air resistance is not considered on assuming that it is negligible.

Explanation of Solution

Note that in any of the calculations given, the influence of air drag is not considered. If taken, calculation will be more complex. Air drag is directly proportional to the square of velocity of projectile. Since the velocity of projectile is small, it can be assumed that air drag has no significant role in the projectile motion of seed.

Thus, the role of air resistance is not considered on assuming that it is negligible.