Chapter 5, Problem 14A

The vectors represent initial velocities of projectiles launched at ground level.

a. Rank them by their vertical components of velocity from greatest to least.

b. Rank them by their horizontal components of velocity from greatest to least.

To determine

To rank:The initial velocities of projectile by their vertical componentsfrom greatest to least.

Answer to Problem 14A

The rank of the initial velocities of projectile by their vertical components of velocities from greatest to least is $B>C>A>D$ .

Explanation of Solution

Given:

The given vectors are shown below.

  

Formula used:

The expression for the vertical component of velocity is

  $v_{\text{y}}=v\sin \theta$

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

Calculation:

The vertical component of velocity for case A is,

  $\begin{array}{l}{v}_{\text{y}}=v\sin \theta \\ v_{\text{Ay}}=\left(10\text{m}/\text{s}\right)\sin 37°\\ v_{\text{Ay}}=6\text{m}/\text{s}\end{array}$

The vertical component of velocity for case B is,

  $\begin{array}{l}{v}_{\text{y}}=v\sin \theta \\ v_{\text{By}}=\left(20\text{m}/\text{s}\right)\sin 53°\\ v_{\text{By}}=16\text{m}/\text{s}\end{array}$

The vertical component of velocity for case C is,

  $\begin{array}{l}{v}_{\text{y}}=v\sin \theta \\ v_{\text{Cy}}=\left(10\text{m}/\text{s}\right)\sin 45°\\ v_{\text{Cy}}=7\text{m}/\text{s}\end{array}$

The vertical component of velocity for case D is,

  $\begin{array}{l}{v}_{\text{y}}=v\sin \theta \\ v_{\text{Dy}}=\left(10\text{m}/\text{s}\right)\sin 30°\\ v_{\text{Dy}}=5\text{m}/\text{s}\end{array}$

Hence, the rank of the initial velocities of projectile by their vertical components of velocities from greatest to least is $B>C>A>D$ .

Conclusion:

Thus, the rank of the initial velocities of projectile by their vertical components of velocities from greatest to least is $B>C>A>D$ .

To determine

To rank:The initial velocities of projectile by their horizontal componentsfrom greatest to least.

Answer to Problem 14A

The rank of the initial velocities of projectile by their horizontal components of velocities from greatest to least is $B>D>A>C$ .

Explanation of Solution

Given:

The given vectors are shown below.

  

Formula used:

The expression for the horizontal component of velocity is,

  $v_{\text{x}}=v\cos \theta$

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

Calculation:

The horizontal component of velocity for case A is,

  $\begin{array}{l}{v}_{\text{x}}=v\cos \theta \\ v_{\text{Ax}}=\left(10\text{m}/\text{s}\right)\cos 37°\\ v_{\text{Ax}}=8\text{m}/\text{s}\end{array}$

The horizontal component of velocity for case B is,

  $\begin{array}{l}{v}_{\text{x}}=v\cos \theta \\ v_{\text{Bx}}=\left(20\text{m}/\text{s}\right)\cos 53°\\ v_{\text{Bx}}=12\text{m}/\text{s}\end{array}$

The horizontal component of velocity for case C is,

  $\begin{array}{l}{v}_{\text{x}}=v\cos \theta \\ v_{\text{Cx}}=\left(10\text{m}/\text{s}\right)\cos 45°\\ v_{\text{Cx}}=7\text{m}/\text{s}\end{array}$

The horizontal component of velocity for case D is,

  $\begin{array}{l}{v}_{\text{x}}=v\cos \theta \\ v_{\text{Dx}}=\left(10\text{m}/\text{s}\right)\cos 30°\\ v_{\text{Dx}}=9\text{m}/\text{s}\end{array}$

Hence, the rank of the initial velocities of projectile by their horizontal components of velocities from greatest to least is $B>D>A>C$ .

Conclusion:

Thus, the rank of the initial velocities of projectile by their horizontal components of velocities from greatest to least is $B>D>A>C$ .