Q53P - University Physics with Modern Physics

Question

According to Guinness World Records, the longest home run ever measured was hit by Roy “Dizzy” Carlyle in a minor league game. The ball traveled $\mathbf{188}\mathbf{ }\mathbf{m}\left(618 \mathrm{ft}\right)$ before landing on the ground outside the ballpark. (a) If the ball’s initial velocity was in a direction ${\mathbf{45}}^{\mathbf{o}}$ above the horizontal, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point $\mathbf{0}\mathbf{.}\mathbf{9}\mathbf{m}\mathbf{ }\left(3.0\mathrm{ft}\right)$ above ground level? Ignore air resistance, and assume that the ground was perfectly flat. (b) How far would the ball be above a fence $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{ }\left(10\mathrm{ft}\right)$ high if the fence was $\mathbf{116}\mathbf{m}\mathbf{ }\left(380\mathrm{ft}\right)$ from home plate?

Step-by-Step Solution

Answer

1. The initial velocity will be $43.0 \mathrm{m}/\mathrm{s}$.
2. The height of the ball above the fence $41.768 \mathrm{m}$.

Step 1: Newton’s law of motion:

Newton’s laws of motion show the relationship between the moving particle or object’s displacement, initial and final velocity, acceleration, and time.

 

The velocity contains two components, one is a horizontal component and another one is a vertical. 

 

According to Newton’s laws of motion,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ 

 

Here, s, u, a, and t are displacement, initial velocity, time, and acceleration respectively.

Step 2: Consider the given data:

Horizontal distance traveled by the ball, ${\mathrm{s}}_{\mathrm{Hb}}=188 \mathrm{m}$  

 

Angle, $\mathrm{\theta }={45}^{\mathrm{o}}$ 

 

Height of the ball, ${\mathrm{s}}_{\mathrm{v}}=0.9 \mathrm{m}$  

 

Height of the fence, $\mathrm{h}=3.0 \mathrm{m}$ 

 

Acceleration due to gravity, $\mathrm{g}=9.8 \mathrm{m}/{\mathrm{s}}^2$ 

 

The horizontal distance of the fence, ${\mathrm{s}}_{\mathrm{Hf}}=116 \mathrm{m}$ 

Step 3: (a) Initial velocity:

For the horizontal motion the gravity is zero. Therefore,

$\begin{array}{rcl}{\mathrm{s}}_{\mathrm{Hb}}& =& \mathrm{utcos\theta }+\frac{1}{2}{\mathrm{gt}}^2\\ & =& \mathrm{utcos\theta }+0\end{array}$ 

$\begin{array}{rcl}\mathrm{t}& =& \frac{{\mathrm{s}}_{\mathrm{Hb}}}{\mathrm{ucos\theta }}\\ & =& \frac{188 \mathrm{m}}{\mathrm{ucos} {45}^{\mathrm{o}}}\end{array}$ 

                                                                  

For the vertical motion, gravity is negative. Therefore, the equation of motion will becomes,

$\begin{array}{rcl}{\mathrm{s}}_{\mathrm{v}}& =& \mathrm{usin\theta t}-\frac{1}{2}{\mathrm{gt}}^2\\ -0.9 \mathrm{m}& =& \mathrm{u} \sin {45}^{\mathrm{o}}\left(\frac{188 \mathrm{m}}{\mathrm{u} \cos {45}^{\mathrm{o}}}\right)-\frac{1}{2}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{188 \mathrm{m}}{\mathrm{u} \cos {45}^{\mathrm{o}}}\right)}^2\\ -0.9 \mathrm{m}& =& 188 \mathrm{m}-\left(\frac{3.4647\times {10}^5}{{\mathrm{u}}^2}\right)\\ 187.1& =& \frac{3.4647\times {10}^5}{{\mathrm{u}}^2}\\ {\mathrm{u}}^2& =& 0.185\times {10}^4\\ \mathrm{u}& =& 43.0 \mathrm{m}/\mathrm{s}\end{array}$ 

 

 

Hence, the initial velocity is 43.0 m/s.

Step 4: (b) Distance of ball from the fence:

For the horizontal motion, gravity is zero

$\begin{array}{rcl}{\mathrm{s}}_{\mathrm{Hf}}& =& \mathrm{ucos\theta }\times \mathrm{t}+\frac{1}{2}{\mathrm{gt}}^2\\ & =& \mathrm{ucos}{45}^{\mathrm{o}}\times \mathrm{t}+0\end{array}$ 

$\begin{array}{rcl}\mathrm{t}& =& \frac{{\mathrm{s}}_{\mathrm{Hf}}}{\mathrm{u} \cos {45}^{\mathrm{o}}}\\ & =& \frac{116 \mathrm{m}}{43.0 \mathrm{m}/\mathrm{s}\times 0.707}\\ & =& 3.82 \mathrm{s}\end{array}$ 

 For the vertical motion, gravity is negative

 $\begin{array}{rcl}\mathrm{s}& =& \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 30.3 \mathrm{m}/\mathrm{s}\times 3.82 \mathrm{s}+\frac{1}{2}\times \left(-9.8 \mathrm{m}/{\mathrm{s}}^2\right)\times {\left(3.83 \mathrm{s}\right)}^2\\ & =& 115.746 \mathrm{m}-71.878 \mathrm{m}\\ & =& 43.87 \mathrm{m}\end{array}$

Hence, the height of ball is 0.90 m and the height of fence is 3.0 m . Now the height of the ball above the fence will be,

$43.87 \mathrm{m}+0.90 \mathrm{m}-3.0 \mathrm{m}=41.768 \mathrm{m}$