Chapter 11.2, Problem 33E

(a)

To determine

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

Answer to Problem 33E

The required position vector is $\left(51.621t\text{feet},-16t^2+73.724t\text{feet}\right)$ .

Explanation of Solution

Given information:

The speed of the ball = 90 feet per second

Angle made by the ball from the horizontal traveling = $55°$

Boundary of the fence = 20 feet

Distance of the fence from the batter = 230 feet

Calculation:

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

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

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(90\right)\cos \left(55°\right)\\ {\overrightarrow{v}}_x=51.621\end{array}$

So, the x -component of the vector is 51.621 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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Hence, the required position vector is $\left(51.621t\text{feet},-16t^2+73.724t\text{feet}\right)$ .

(b)

To determine

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

Answer to Problem 33E

The required velocity vector is $\left(51.621,-32t+73.724\right)$ .

Explanation of Solution

Given information:

The speed of the ball = 90 feet per second

Angle made by the ball from the horizontal traveling = $55°$

Boundary of the fence = 20 feet

Distance of the fence from the batter = 230 feet

Calculation:

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

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

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(90\right)\cos \left(55°\right)\\ {\overrightarrow{v}}_x=51.621\end{array}$

So, the x -component of the vector is 51.621 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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Thus, the position vector is $\left(51.621t\text{feet},-16t^2+73.724t\text{feet}\right)$ .

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

So, $\begin{array}{c}\overrightarrow{V}=\left(\frac{d}{dt}\left(51.621t\right),\frac{d}{dt}\left(-{16}^2+73.724t\right)\right)\\ =\left(51.621,-32t+73.724\right)\end{array}$

Hence, the required velocity vector is $\left(51.621,-32t+73.724\right)$ .

(c)

To determine

To determine: Whether the hit is a home run or not.

Answer to Problem 33E

The hit is not a home run.

Explanation of Solution

Given information:

The speed of the ball = 90 feet per second

Angle made by the ball from the horizontal traveling = $55°$

Boundary of the fence = 20 feet

Distance of the fence from the batter = 230 feet

Calculation:

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

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

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(90\right)\cos \left(55°\right)\\ {\overrightarrow{v}}_x=51.621\end{array}$

So, the x -component of the vector is 51.621 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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Thus, the position vector is $\left(51.621\text{feet},-16t^2+73.724t\text{feet}\right)$ .

To find whether the hit is a home run, use the y -component of the position vector to see whether it is greater than 20 feet.

Also, consider that $x=230$ .

Substitute 230 for x in the expression $51.621t=x$ and solve for t .

  $\begin{array}{c}51.621t=230\\ t=\frac{230}{51.621}\\ t\approx 4.456\end{array}$

Substitute 4.456 for t in the expression $y=-16t^2+73.724t$ .

  $\begin{array}{c}y=-16{\left(4.456\right)}^2+73.724\left(4.456\right)\\ y=10.819\end{array}$

It is seen that the y -component not greater than 20 feet.

Hence, the hit is not a home run.

(d)

To determine

To determine: The time at which the ball hits the fence or clears it for a home run.

Answer to Problem 33E

The required time at which the ball hits the fence or clears it for a home run is approximately 4.456 seconds.

Explanation of Solution

Given information:

The speed of the ball = 90 feet per second

Angle made by the ball from the horizontal traveling = $55°$

Boundary of the fence = 20 feet

Distance of the fence from the batter = 230 feet

Calculation:

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

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

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(90\right)\cos \left(55°\right)\\ {\overrightarrow{v}}_x=51.621\end{array}$

So, the x -component of the vector is 51.621 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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Thus, the position vector is $\left(51.621\text{feet},-16t^2+73.724t\text{feet}\right)$ .

To find the time at which the ball hits the fence or clears it for a home run, use the x -component of the position vector where $x=230$ .

Substitute 230 for x in the expression $51.621t=x$ and solve for t .

  $\begin{array}{c}51.621t=230\\ t=\frac{230}{51.621}\\ t\approx 4.456\end{array}$

Hence, the required time at which the ball hits the fence or clears it for a home run is approximately 4.456 seconds.

(e)

To determine

To determine: The velocity of the ball when it either hits the fence or passes over it.

Answer to Problem 33E

The required velocity of the ball is 86.06 feet per second.

Explanation of Solution

Given information:

The speed of the ball = 90 feet per second

Angle made by the ball from the horizontal traveling = $55°$

Boundary of the fence = 20 feet

Distance of the fence from the batter = 230 feet

Calculation:

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

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

The horizontal component of the velocity is:

  $\begin{array}{l}{\overrightarrow{v}}_x=\overrightarrow{v}\cos \theta \\ {\overrightarrow{v}}_x=\left(90\right)\cos \left(55°\right)\\ {\overrightarrow{v}}_x=51.621\end{array}$

So, the x -component of the vector is 51.621 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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Thus, the position vector is $\left(51.621\text{feet},-16t^2+73.724t\text{feet}\right)$ .

To find the time at which the ball hits the fence or clears it for a home run, use the x -component of the position vector where $x=230$ .

Substitute 230 for x in the expression $51.621t=x$ and solve for t .

  $\begin{array}{c}51.621t=230\\ t=\frac{230}{51.621}\\ t\approx 4.456\end{array}$

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:

  $\begin{array}{l}b={\overrightarrow{v}}_y\\ {\overrightarrow{v}}_y=\overrightarrow{v}\sin \theta \\ {\overrightarrow{v}}_y=\left(90\right)\sin \left(55°\right)\\ {\overrightarrow{v}}_y=73.724\end{array}$

Substitute 73.724 for b in the equation $y=-16t^2+bt$ .

  $y=-16t^2+73.724t$

Thus, the position vector is $\left(51.621\text{feet},-16t^2+73.724t\text{feet}\right)$ .

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

  $\begin{array}{c}\overrightarrow{V}=\left(\frac{d}{dt}\left(51.621t\right),\frac{d}{dt}\left(-{16}^2+73.724t\right)\right)\\ =\left(51.621,-32t+73.724\right)\end{array}$

Thus, the velocity vector is $\left(51.621,-32t+73.724\right)$ .

Now, to find the velocity of the ball find the magnitude of the velocity vector using the formula $\overrightarrow{v}=\left|\langle \overrightarrow{x},\overrightarrow{y}\rangle \right|=\sqrt{{\overrightarrow{x}}^2+{\overrightarrow{y}}^2}$ .

Substitute 4.456 for t in the velocity vector $\overrightarrow{V}=\left(51.621,-32t+73.724\right)$ .

  $\begin{array}{l}\overrightarrow{V}=\left(51.621,-32\left(4.456\right)+73.724\right)\\ \overrightarrow{V}=\left(51.621,68.86\right)\end{array}$

Again, substitute 51.621 for $\overrightarrow{x}$ and 68.86 for $\overrightarrow{y}$ in the formula $\overrightarrow{v}=\left|\langle \overrightarrow{x},\overrightarrow{y}\rangle \right|=\sqrt{{\overrightarrow{x}}^2+{\overrightarrow{y}}^2}$ .

  $\begin{array}{l}\overrightarrow{v}=\sqrt{{\left(51.621\right)}^2+{\left(68.86\right)}^2}\\ \overrightarrow{v}=\sqrt{7406.3236}\\ \overrightarrow{v}=86.06\end{array}$

Hence, the required velocity of the ball is 86.06 feet per second.