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.

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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.
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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