ou want to land a projectile 1 m away from where it is launched at 4.4 m/s. What gles can be used?

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Chapter1: Units, Trigonometry. And Vectors
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I need help with question 8

**Problem 8: Projectile Motion**

You want to land a projectile 1 meter away from where it is launched at a speed of 4.4 meters per second. What angles can be used?

**Explanation:**

This problem involves determining the launch angles required to land a projectile exactly 1 meter away, given a launch speed of 4.4 m/s. The solution involves applying principles of projectile motion, specifically calculating the range of a projectile using its initial velocity and launch angle. 

To solve the problem:

1. **Use the range formula for projectile motion**, which is given by:

   \[
   R = \frac{v^2 \sin(2\theta)}{g}
   \]

   Where:
   - \( R \) is the range (1 meter in this problem),
   - \( v \) is the initial velocity (4.4 m/s),
   - \( \theta \) is the launch angle,
   - \( g \) is the acceleration due to gravity (approximately 9.81 m/s\(^2\)).

2. **Solve for the angle \(\theta\)**, knowing that the range \( R \) should equal 1 meter. 

By substituting the known values into the range equation, students can solve for the angle(s) \(\theta\) that will result in the projectile traveling exactly 1 meter. Graphics or diagrams depicting the trajectory of the projectile would show the initial velocity vector, the height and range, and how varying \(\theta\) changes the path.
Transcribed Image Text:**Problem 8: Projectile Motion** You want to land a projectile 1 meter away from where it is launched at a speed of 4.4 meters per second. What angles can be used? **Explanation:** This problem involves determining the launch angles required to land a projectile exactly 1 meter away, given a launch speed of 4.4 m/s. The solution involves applying principles of projectile motion, specifically calculating the range of a projectile using its initial velocity and launch angle. To solve the problem: 1. **Use the range formula for projectile motion**, which is given by: \[ R = \frac{v^2 \sin(2\theta)}{g} \] Where: - \( R \) is the range (1 meter in this problem), - \( v \) is the initial velocity (4.4 m/s), - \( \theta \) is the launch angle, - \( g \) is the acceleration due to gravity (approximately 9.81 m/s\(^2\)). 2. **Solve for the angle \(\theta\)**, knowing that the range \( R \) should equal 1 meter. By substituting the known values into the range equation, students can solve for the angle(s) \(\theta\) that will result in the projectile traveling exactly 1 meter. Graphics or diagrams depicting the trajectory of the projectile would show the initial velocity vector, the height and range, and how varying \(\theta\) changes the path.
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Motion of projectile can be divided into x component and y component. 

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