Help with # 5 Certainly! Here's a transcription and explanation suitable for an educational website:---**Horizontally Launched Projectiles****Problem 5:** Will Smith pushes the iRobot over the ledge to its doom. If the robot had a horizontal velocity of 2.3 m/s and was 17 m above the ground, what was the robot’s horizontal displacement when it hit the ground?---**Part B Instructions:** Use the information above to solve each of the following problems. Check your answers below. Attach your work to the document.---**Explanation of Concepts:**When an object is launched horizontally, its motion can be analyzed in two dimensions: horizontal and vertical. The horizontal motion occurs at constant velocity since there is no acceleration (ignoring air resistance), while the vertical motion is influenced by gravity.1. **Horizontal Motion:** - **Initial Velocity ($v_x$)**: 2.3 m/s - The horizontal displacement ($d_x$) can be calculated using the formula $d_x = v_x \times t$, where $t$ is the time of flight.2. **Vertical Motion:** - **Initial Position ($y_0$)**: 17 m above the ground - The vertical displacement and time of flight can be calculated using the formula $y = y_0 - \frac{1}{2} g t^2$, where $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity).Use these equations to calculate the time it takes for the robot to reach the ground and subsequently find the horizontal displacement.

Answered: 5. Will smith pushes the irobot over…

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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Help with # 5

Certainly! Here's a transcription and explanation suitable for an educational website:

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**Horizontally Launched Projectiles**

**Problem 5:**
Will Smith pushes the iRobot over the ledge to its doom. If the robot had a horizontal velocity of 2.3 m/s and was 17 m above the ground, what was the robot’s horizontal displacement when it hit the ground?

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**Part B Instructions:**
Use the information above to solve each of the following problems. Check your answers below. Attach your work to the document.

---

**Explanation of Concepts:**

When an object is launched horizontally, its motion can be analyzed in two dimensions: horizontal and vertical. The horizontal motion occurs at constant velocity since there is no acceleration (ignoring air resistance), while the vertical motion is influenced by gravity.

1. **Horizontal Motion:**
- **Initial Velocity ($v_x$)**: 2.3 m/s
- The horizontal displacement ($d_x$) can be calculated using the formula $d_x = v_x \times t$, where $t$ is the time of flight.

2. **Vertical Motion:**
- **Initial Position ($y_0$)**: 17 m above the ground
- The vertical displacement and time of flight can be calculated using the formula $y = y_0 - \frac{1}{2} g t^2$, where $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity).

Use these equations to calculate the time it takes for the robot to reach the ground and subsequently find the horizontal displacement.

Expert Solution

Step 1

Given

The horizontal velocity is u_h = 2.3 m/s.

The height from which robot is pushes or the vertical distance cover by the robot is h = 17 m.

By the newton equation of motion, the vertical distance formula is given as,

$h = u_{v} \times t + \frac{1}{2} g t^{2}$

Here, u_v is the vertical initial velocity which is equal to 0 as it is free fall from the height, t is the time it hits the ground and g is acceleration due to gravity whose value is 9.8 m/s².

Substitute the known values.

$\begin{array}{rcl} 17 m & = & 0 \times t + \frac{1}{2} \times 9.8 m / s^{2} \times t^{2} \\ t^{2} & = & \frac{17 m \times 2}{9.8 m / s^{2}} \\ = & 3.4694 s^{2} \\ t & = & \sqrt{3.4694 s^{2}} \\ = & 1.86 s \end{array}$

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