A baseball is thrown toward a player with an initial speed of 36.6 m/s at 50⁰ with the horizontal. At the moment the ball is thrown, the player is 150 m from the thrower. At what average speed must he run toward the thrower to catch the baseball 0.914 m above the height at which it was released?

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Chapter1: Units, Trigonometry. And Vectors
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**Projectile Motion Problem**

**Problem Statement:**

A baseball is thrown toward a player with an initial speed of 36.6 m/s at an angle of 50° with the horizontal. At the moment the ball is thrown, the player is 150 m from the thrower. At what average speed must the player run toward the thrower to catch the baseball 0.914 m above the height at which it was released?

**Solution Approach:**

1. **Break Down the Initial Velocity:**
   - Determine the horizontal (vx) and vertical (vy) components of the initial velocity (v0).
     \[ v_0 x = v_0 \cos(\theta) \]
     \[ v_0 y = v_0 \sin(\theta) \]
     Where \( v_0 = 36.6 \) m/s and \( \theta = 50° \).

2. **Calculate the Time of Flight:**
   - Use concepts from projectile motion to determine the time it takes for the ball to reach the catching height, 0.914 m above the release point.

3. **Horizontal Motion Analysis:**
   - Calculate the horizontal distance the ball travels during this time.
   - Determine the relative motion between the player and the ball.

4. **Player's Required Speed:**
   - Establish the player's required average speed to intercept the ball at the specified location.

The key aspect is to correctly compute the dynamics involving both vertical and horizontal components of motion due to gravity, and then use this information to find the player's running speed.

By solving the above steps one can find the average speed at which the player must run toward the thrower to successfully catch the baseball.
Transcribed Image Text:**Projectile Motion Problem** **Problem Statement:** A baseball is thrown toward a player with an initial speed of 36.6 m/s at an angle of 50° with the horizontal. At the moment the ball is thrown, the player is 150 m from the thrower. At what average speed must the player run toward the thrower to catch the baseball 0.914 m above the height at which it was released? **Solution Approach:** 1. **Break Down the Initial Velocity:** - Determine the horizontal (vx) and vertical (vy) components of the initial velocity (v0). \[ v_0 x = v_0 \cos(\theta) \] \[ v_0 y = v_0 \sin(\theta) \] Where \( v_0 = 36.6 \) m/s and \( \theta = 50° \). 2. **Calculate the Time of Flight:** - Use concepts from projectile motion to determine the time it takes for the ball to reach the catching height, 0.914 m above the release point. 3. **Horizontal Motion Analysis:** - Calculate the horizontal distance the ball travels during this time. - Determine the relative motion between the player and the ball. 4. **Player's Required Speed:** - Establish the player's required average speed to intercept the ball at the specified location. The key aspect is to correctly compute the dynamics involving both vertical and horizontal components of motion due to gravity, and then use this information to find the player's running speed. By solving the above steps one can find the average speed at which the player must run toward the thrower to successfully catch the baseball.
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