**Projectile Motion Problem**A projectile is fired from the ground level at an initial angle of \(80^\circ\) above the horizontal and travels 14 meters away (horizontal distance) from its starting position. (\(t\) is measured in seconds.)**(a) Initial Speed Calculation**Calculate its **initial speed**. Round your answer to the **nearest tenth**.*Detailed Calculation:*1. First, resolve the given angle and initial speed into horizontal and vertical components. 2. Use the formula for horizontal range \(R\):\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]3. Where: - \( R \) is the range (14 meters), - \( \theta \) is the angle of projection (80 degrees), - \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)), - \( v_0 \) is the initial speed.4. Rearrange the formula to solve for \( v_0 \):\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \]5. Plug in the known values and calculate.**(b) Maximum Height Calculation**Calculate its **maximum height** above the ground level. Round your answer to the **nearest tenth**.*Detailed Calculation:*1. Use the vertical component of the initial velocity \(v_{0y}\):\[ v_{0y} = v_0 \sin(\theta) \]2. Use the formula for maximum height \(H\):\[ H = \frac{v_{0y}^2}{2g} \]3. Where \( v_{0y} = v_0 \sin(\theta) \).4. Substitute the initial speed found in part (a) and solve for \( H \).

Name: **Projectile Motion Problem**A projectile is fired from the ground le
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Description: **Projectile Motion Problem**A projectile is fired from the ground level at an initial angle of \(80^\circ\) above the horizontal and travels 14 meters away (horizontal distance) from its starting position. (\(t\) is measured in seconds.)**(a) Initi

As we know, θ = 80° (a) Range (R) = u^2 Sin 2θ/g R = 14 m u^2 = 401.15 u = 20.03 m/s

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A projectile is fired from the ground level at an initial angle of 80' above the horizontal and travels 14 meters away (horizontal distance) from its starting position. (t is measured in seconds.) (a) Calculate its initial speed. Round your answer to the nearest tenth. (b) Calculate its maximum height above the ground level. Round your answer to the nearest tenth.

A projectile is fired from the ground level at an initial angle of 80' above the horizontal and travels 14 meters away (horizontal distance) from its starting position. (t is measured in seconds.) (a) Calculate its initial speed. Round your answer to the nearest tenth. (b) Calculate its maximum height above the ground level. Round your answer to the nearest tenth.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Projectile Motion

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**Projectile Motion Problem**

A projectile is fired from the ground level at an initial angle of \(80^\circ\) above the horizontal and travels 14 meters away (horizontal distance) from its starting position. (\(t\) is measured in seconds.)

**(a) Initial Speed Calculation**

Calculate its **initial speed**. Round your answer to the **nearest tenth**.

*Detailed Calculation:*

1. First, resolve the given angle and initial speed into horizontal and vertical components.
2. Use the formula for horizontal range \(R\):

\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]

3. Where:
- \( R \) is the range (14 meters),
- \( \theta \) is the angle of projection (80 degrees),
- \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)),
- \( v_0 \) is the initial speed.

4. Rearrange the formula to solve for \( v_0 \):

\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \]

5. Plug in the known values and calculate.

**(b) Maximum Height Calculation**

Calculate its **maximum height** above the ground level. Round your answer to the **nearest tenth**.

*Detailed Calculation:*

1. Use the vertical component of the initial velocity \(v_{0y}\):

\[ v_{0y} = v_0 \sin(\theta) \]

2. Use the formula for maximum height \(H\):

\[ H = \frac{v_{0y}^2}{2g} \]

3. Where \( v_{0y} = v_0 \sin(\theta) \).

4. Substitute the initial speed found in part (a) and solve for \( H \).

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