A ball thrown straight up into the air is found to be moving at 7.79 m/s after falling 2.95 m below its release point. Find the ball's initlal speed (in m/s). HINT 10.89 X m/s

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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement:
A ball thrown straight up into the air is found to be moving at 7.79 m/s after falling 2.95 m below its release point. Find the ball’s initial speed (in m/s).

### Problem Input Interface:
- **Text Box**: For entering the initial speed of the ball (in m/s).
- **Hint Button**: Provides a hint if clicked.
- **Need Help?**: Two buttons:
  - **Read It Button**: Presumably opens a guide or textbook reference.
  - **Watch It Button**: Presumably plays a video tutorial on solving a similar problem.

### Example Input and Feedback:
- **Entered Answer**: 10.89 m/s
- **Feedback**: Marked as incorrect (indicated by a red "X" next to the answer).

### Solve the Problem:
To solve for the initial speed \(v_0\) of the ball, one can use the kinematic equations of motion, under the influence of gravity. The process involves considering the principles of energy or the kinematic equations that relate velocity, acceleration, distance, and time.

1. **Kinematic Equation Approach**:
   - Given data:
     - Final speed \(v = 7.79\) m/s (downwards, so we take it as negative)
     - Distance fallen \(s = 2.95\) m (downwards)
     - Acceleration due to gravity \(g = 9.81\) m/s² (downwards, should be taken as positive)

   - The kinematic equation:
     \[
     v^2 = v_0^2 + 2as
     \]
     Where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.
     
   - Substituting the known values:
     \[
     (-7.79)^2 = v_0^2 + 2(9.81)(-2.95)
     \]
     Simplifying:
     \[
     60.76 = v_0^2 - 57.89
     \]
     \[
     v_0^2 = 60.76 + 57.89
     \]
     \[
     v_0^2 = 118.65
     \]
     \[
Transcribed Image Text:### Problem Statement: A ball thrown straight up into the air is found to be moving at 7.79 m/s after falling 2.95 m below its release point. Find the ball’s initial speed (in m/s). ### Problem Input Interface: - **Text Box**: For entering the initial speed of the ball (in m/s). - **Hint Button**: Provides a hint if clicked. - **Need Help?**: Two buttons: - **Read It Button**: Presumably opens a guide or textbook reference. - **Watch It Button**: Presumably plays a video tutorial on solving a similar problem. ### Example Input and Feedback: - **Entered Answer**: 10.89 m/s - **Feedback**: Marked as incorrect (indicated by a red "X" next to the answer). ### Solve the Problem: To solve for the initial speed \(v_0\) of the ball, one can use the kinematic equations of motion, under the influence of gravity. The process involves considering the principles of energy or the kinematic equations that relate velocity, acceleration, distance, and time. 1. **Kinematic Equation Approach**: - Given data: - Final speed \(v = 7.79\) m/s (downwards, so we take it as negative) - Distance fallen \(s = 2.95\) m (downwards) - Acceleration due to gravity \(g = 9.81\) m/s² (downwards, should be taken as positive) - The kinematic equation: \[ v^2 = v_0^2 + 2as \] Where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement. - Substituting the known values: \[ (-7.79)^2 = v_0^2 + 2(9.81)(-2.95) \] Simplifying: \[ 60.76 = v_0^2 - 57.89 \] \[ v_0^2 = 60.76 + 57.89 \] \[ v_0^2 = 118.65 \] \[
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