As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 21.5 m/s. In the process, he moves his hand through a distance of 1.45 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.
As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 21.5 m/s. In the process, he moves his hand through a distance of 1.45 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.
Physics for Scientists and Engineers: Foundations and Connections
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ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter2: One Dimensional Motion
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![**Problem Statement:**
As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 21.5 m/s. In the process, he moves his hand through a distance of 1.45 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.
**Variables Given:**
- Initial speed (v) = 21.5 m/s
- Distance (d) = 1.45 m
- Mass (m) = 0.150 kg
**Objective:**
Calculate the force exerted on the ball.
**Calculation:**
To solve this problem, you would typically use the work-energy principle which relates the work done on the ball to its change in kinetic energy. The work done (W) by the force (F) is given by:
\[ W = F \cdot d \]
The change in kinetic energy (ΔKE) is given by:
\[ ΔKE = \frac{1}{2} m v^2 - \frac{1}{2} m \cdot 0^2 = \frac{1}{2} m v^2 \]
Since the work done is equal to the change in kinetic energy:
\[ F \cdot d = \frac{1}{2} m v^2 \]
Solving for F (Force):
\[ F = \frac{\frac{1}{2} m v^2}{d} \]
**Graphs and Diagrams:**
No graphs or diagrams are present in the image. This explanation would typically accompany a diagram showing a pitcher throwing the ball, with forces labeled and an indication of the initial velocity and distance moved by the hand.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F212f1a9d-c23e-4230-9cf0-4b21895c9148%2Fa2d8a706-155e-4a6b-bcff-4f4fd425e06c%2Foaifyo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 21.5 m/s. In the process, he moves his hand through a distance of 1.45 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.
**Variables Given:**
- Initial speed (v) = 21.5 m/s
- Distance (d) = 1.45 m
- Mass (m) = 0.150 kg
**Objective:**
Calculate the force exerted on the ball.
**Calculation:**
To solve this problem, you would typically use the work-energy principle which relates the work done on the ball to its change in kinetic energy. The work done (W) by the force (F) is given by:
\[ W = F \cdot d \]
The change in kinetic energy (ΔKE) is given by:
\[ ΔKE = \frac{1}{2} m v^2 - \frac{1}{2} m \cdot 0^2 = \frac{1}{2} m v^2 \]
Since the work done is equal to the change in kinetic energy:
\[ F \cdot d = \frac{1}{2} m v^2 \]
Solving for F (Force):
\[ F = \frac{\frac{1}{2} m v^2}{d} \]
**Graphs and Diagrams:**
No graphs or diagrams are present in the image. This explanation would typically accompany a diagram showing a pitcher throwing the ball, with forces labeled and an indication of the initial velocity and distance moved by the hand.
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