Chapter 11.2, Problem 34E

(a)

To determine

To determine: The position vector of the football at time t seconds.

Answer to Problem 34E

The required position vector is $\left(90+44.116t\text{feet},-16t^2+67.932t\text{feet}\right)$ .

Explanation of Solution

Given information:

Length of the player’s own line = 30 yards

Angle made by the ball from horizontal = $57°$

Speed of the ball = 81 feet per second

Height of the player of the opposing team = 6 feet

Length of the 6-foot player’s line = 10 yard

Calculation:

Consider that the air resistance is negligible while the ball leaves the player.

First, find the horizontal position of the ball using the cosine function.

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(81\right)\cos \left(57°\right)\\ {\overrightarrow{v}}_x=44.116\end{array}$

So, the x -component of the vector is $90+44.116t$ feet.

The formula for position of a particle with initial velocity and initial height is $y=-16t^2+bt+c$ .

Since, the initial height is 0;

Substitute 0 for c in the above formula.

  $\begin{array}{c}y=-16t^2+bt+0\\ y=-16t^2+bt\end{array}$

Here, b is the initial velocity which is ${\overrightarrow{v}}_y$ .

So, $b={\overrightarrow{v}}_y$

The vertical position of the ball can be calculated using the sine function.

The vertical component of the velocity is:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(81\right)\sin \left(57°\right)\\ {\overrightarrow{v}}_y=67.932\end{array}$

Substitute 67.932 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+67.932t$

Hence, the required position vector is $\left(90+44.116t\text{feet},-16t^2+67.932t\text{feet}\right)$ .

(b)

To determine

To determine: The velocity vector of the baseball at time t seconds.

Answer to Problem 34E

The required velocity vector is $\left(44.116,-8t+67.932\right)$ .

Explanation of Solution

Given information:

Length of the player’s own line = 30 yards

Angle made by the ball from horizontal = $57°$

Speed of the ball = 81 feet per second

Height of the player of the opposing team = 6 feet

Length of the 6-foot player’s line = 10 yard

Calculation:

Consider that the air resistance is negligible while the ball leaves the player.

First, find the horizontal position of the ball using the cosine function.

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(81\right)\cos \left(57°\right)\\ {\overrightarrow{v}}_x=44.116\end{array}$

So, the x -component of the vector is $90+44.116t$ feet.

The formula for position of a particle with initial velocity and initial height is $y=-16t^2+bt+c$ .

Since, the initial height is 0;

Substitute 0 for c in the above formula.

  $\begin{array}{c}y=-16t^2+bt+0\\ y=-16t^2+bt\end{array}$

Here, b is the initial velocity which is ${\overrightarrow{v}}_y$ .

So, $b={\overrightarrow{v}}_y$

The vertical position of the ball can be calculated using the sine function.

The vertical component of the velocity is:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(81\right)\sin \left(57°\right)\\ {\overrightarrow{v}}_y=67.932\end{array}$

Substitute 67.932 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+67.932t$

Thus, the position vector is $\left(90+44.116t\text{feet},-16t^2+67.932t\text{feet}\right)$ .

To find the velocity vector of the football, find the derivative of each of the components of the above position vector.

So, $\begin{array}{l}\overrightarrow{V}=\left(\frac{d}{dt}\left(90+44.116t\right),\frac{d}{dt}\left(-16t^2+67.932t\right)\right)\\ \overrightarrow{V}=\left(44.116,-8t+67.932\right)\end{array}$

Hence, the required velocity vector is $\left(44.116,-8t+67.932\right)$ .

(c)

To determine

To determine: The time at which the football is over the player on the 10-yard line.

Answer to Problem 34E

The required time at which the football is over the player on the 10-yard line is 4.08 seconds.

Explanation of Solution

Given information:

Length of the player’s own line = 30 yards

Angle made by the ball from horizontal = $57°$

Speed of the ball = 81 feet per second

Height of the player of the opposing team = 6 feet

Length of the 6-foot player’s line = 10 yard

Calculation:

Consider that the air resistance is negligible while the ball leaves the player.

First, find the horizontal position of the ball using the cosine function.

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(81\right)\cos \left(57°\right)\\ {\overrightarrow{v}}_x=44.116\end{array}$

So, the x -component of the vector is $90+44.116t$ feet.

The formula for position of a particle with initial velocity and initial height is $y=-16t^2+bt+c$ .

Since, the initial height is 0;

Substitute 0 for c in the above formula.

  $\begin{array}{c}y=-16t^2+bt+0\\ y=-16t^2+bt\end{array}$

Here, b is the initial velocity which is ${\overrightarrow{v}}_y$ .

So, $b={\overrightarrow{v}}_y$

The vertical position of the ball can be calculated using the sine function.

The vertical component of the velocity is:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(81\right)\sin \left(57°\right)\\ {\overrightarrow{v}}_y=67.932\end{array}$

Substitute 67.932 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+67.932t$

Thus, the position vector is $\left(90+44.116t\text{feet},-16t^2+67.932t\text{feet}\right)$ .

To find the time at which the football is over the player on the 10-yard line, use the x -component of the position vector where $x=270$ .

Substitute 270 for x in the expression $51.621t=x$ and solve for t .

  $\begin{array}{c}90+44.116t=270\\ \frac{44.116t}{44.116}=\frac{270-90}{44.116}\\ t=\frac{180}{44.116}\\ t=4.08\end{array}$

Hence, the required time at which the football is over the player on the 10-yard line is 4.08 seconds.

(d)

To determine

To determine: Whether the 6-ft player can call for a “fair catch” and likely catch the ball while the standing on the 10-yard line downfield.

Answer to Problem 34E

It is unlikely that the 6-ft player would catch the ball.

Explanation of Solution

Given information:

Length of the player’s own line = 30 yards

Angle made by the ball from horizontal = $57°$

Speed of the ball = 81 feet per second

Height of the player of the opposing team = 6 feet

Length of the 6-foot player’s line = 10 yard

Calculation:

Consider that the air resistance is negligible while the ball leaves the player.

First, find the horizontal position of the ball using the cosine function.

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(81\right)\cos \left(57°\right)\\ {\overrightarrow{v}}_x=44.116\end{array}$

So, the x -component of the vector is $90+44.116t$ feet.

The formula for position of a particle with initial velocity and initial height is $y=-16t^2+bt+c$ .

Since, the initial height is 0;

Substitute 0 for c in the above formula.

  $\begin{array}{c}y=-16t^2+bt+0\\ y=-16t^2+bt\end{array}$

Here, b is the initial velocity which is ${\overrightarrow{v}}_y$ .

So, $b={\overrightarrow{v}}_y$

The vertical position of the ball can be calculated using the sine function.

The vertical component of the velocity is:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(81\right)\sin \left(57°\right)\\ {\overrightarrow{v}}_y=67.932\end{array}$

Substitute 67.932 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+67.932t$

Thus, the position vector is $\left(90+44.116t\text{feet},-16t^2+67.932t\text{feet}\right)$ .

Find the time at which the football is over the player on the 10-yard line using the x -component of the position vector where $x=270$ .

Substitute 270 for x in the expression $51.621t=x$ and solve for t .

  $\begin{array}{c}90+44.116t=270\\ \frac{44.116t}{44.116}=\frac{270-90}{44.116}\\ t=\frac{180}{44.116}\\ t=4.08\end{array}$

Now, to find out whether the 6-ft player can call for a “fair catch” and likely catch the ball while the standing on the 10-yard line downfield, find the height that is, the y -component of the position vector at $t=4.08$ .

Substitute 4.08 for t in the expression $\overrightarrow{y}=-16t^2+67.932t$ and simplify for $\overrightarrow{y}$ .

  $\begin{array}{l}\overrightarrow{y}=-16{\left(4.08\right)}^2+67.932\left(4.08\right)\\ \overrightarrow{y}=14.72\end{array}$

It is seen that the ball is over 14 feet in the air from the 6-ft player.

Hence, it is unlikely that the 6-ft player would catch the ball.