A pitcher throws a 0.140 -kg baseball, and it approaches the bat at a speed of 40.0 m/s. The bat does `W_(nc)= 70.0J` of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.

Options

A. 23.4m/s
B. 48.5m/s
C. 45.9m/s
D. 51.0m/s

Answer

To solve the problem, we will use the work-energy principle, which states that the work done on an object is equal to the change in its mechanical energy (the sum of its kinetic and potential energy).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the baseball, \( m = 0.140 \, \text{kg} \)
- Initial speed of the baseball, \( u = 40.0 \, \text{m/s} \)
- Work done by the bat, \( W_{nc} = 70.0 \, \text{J} \)
- Height after impact, \( h = 25.0 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate Initial Kinetic Energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.140 \times (40.0)^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.140 \times 1600 = 112 \, \text{J} \]

3. **Calculate Potential Energy (PE_final) at height \( h \):**
\[ PE_{\text{final}} = mgh = 0.140 \times 9.8 \times 25.0 \]
\[ PE_{\text{final}} = 0.140 \times 245 = 34.3 \, \text{J} \]

4. **Apply the Work-Energy Principle:**
The work done on the ball is equal to the change in kinetic energy plus the change in potential energy:
\[ W_{nc} = KE_{\text{final}} - KE_{\text{initial}} + PE_{\text{final}} - PE_{\text{initial}} \]
Since the initial potential energy (at the point of impact) is zero:
\[ W_{nc} = KE_{\text{final}} - KE_{\text{initial}} + PE_{\text{final}} \]
Rearranging gives:
\[ KE_{\text{final}} = W_{nc} + KE_{\text{initial}} - PE_{\text{final}} \]

5. **Substitute the Values:**
\[ KE_{\text{final}} = 70.0 + 112 - 34.3 \]
\[ KE_{\text{final}} = 147.7 \, \text{J} \]

6. **Calculate Final Speed (v) using KE_final:**
\[ KE_{\text{final}} = \frac{1}{2} m v^2 \]
\[ 147.7 = \frac{1}{2} \times 0.140 \times v^2 \]
Rearranging gives:
\[ v^2 = \frac{147.7 \times 2}{0.140} \]
\[ v^2 = \frac{295.4}{0.140} = 2103.0 \]
\[ v = \sqrt{2103.0} \approx 45.9 \, \text{m/s} \]

### Final Answer:
The speed of the ball after it leaves the bat and is 25.0 m above the point of impact is approximately **45.9 m/s**.