A baseball is hit into the air at an initial speed of 36.6 m/s and an angle of `50.0^(@)` above the horizontal. At the same time, the center fielder starts running away from the batter, and he catches the ball 0.914 m above the level at which it was hit. If the center fielder is initially `1.10xx10^(2)m` from home plate, what must be his average speed?

Options

A. 3.5 m/s
B. 5.0 m/s
C. 4.2 m/s
D. 6.9 m/s

Answer

To solve the problem, we will break it down into several steps:

### Step 1: Determine the initial velocity components of the baseball.
The initial speed of the baseball is given as \( v_0 = 36.6 \, \text{m/s} \) at an angle of \( \theta = 50^\circ \).

- The horizontal component of the initial velocity \( v_{0x} \) is given by:
\[ v_{0x} = v_0 \cos(\theta) = 36.6 \cos(50^\circ) \]
- The vertical component of the initial velocity \( v_{0y} \) is given by:
\[ v_{0y} = v_0 \sin(\theta) = 36.6 \sin(50^\circ) \]

Calculating these values:
\[ v_{0x} \approx 36.6 \times 0.6428 \approx 23.5 \, \text{m/s} \]
\[ v_{0y} \approx 36.6 \times 0.7660 \approx 28.0 \, \text{m/s} \]

### Step 2: Use the vertical motion equation to find the time of flight until the ball reaches a height of 0.914 m.
Using the second equation of motion:
\[ h = v_{0y} t - \frac{1}{2} g t^2 \]
where \( h = 0.914 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \).

Substituting the values:
\[ 0.914 = 28.0 t - 4.9 t^2 \]
Rearranging gives:
\[ 4.9 t^2 - 28.0 t + 0.914 = 0 \]

### Step 3: Solve the quadratic equation for time \( t \).
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4.9 \), \( b = -28.0 \), and \( c = 0.914 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = (-28.0)^2 - 4 \times 4.9 \times 0.914 \]
\[ D = 784 - 17.912 = 766.088 \]

Now substituting back into the quadratic formula:
\[ t = \frac{28.0 \pm \sqrt{766.088}}{2 \times 4.9} \]
Calculating \( \sqrt{766.088} \approx 27.7 \):
\[ t = \frac{28.0 \pm 27.7}{9.8} \]
Calculating the two possible times:
1. \( t_1 = \frac{55.7}{9.8} \approx 5.68 \, \text{s} \) (valid)
2. \( t_2 = \frac{0.3}{9.8} \approx 0.03 \, \text{s} \) (not valid)

### Step 4: Calculate the horizontal distance traveled by the baseball.
Using the horizontal motion:
\[ d = v_{0x} \cdot t \]
Substituting the values:
\[ d = 23.5 \times 5.68 \approx 133.48 \, \text{m} \]

### Step 5: Determine the distance the center fielder must run.
The center fielder starts at a distance of \( 110 \, \text{m} \) from home plate:
\[ \text{Distance to run} = d - 110 = 133.48 - 110 = 23.48 \, \text{m} \]

### Step 6: Calculate the average speed of the center fielder.
The average speed \( v \) is given by:
\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{23.48}{5.68} \approx 4.13 \, \text{m/s} \]

### Final Answer:
The average speed of the center fielder must be approximately \( 4.13 \, \text{m/s} \).

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