**Physics Lab Experiment: Calculating the Height of Stanley Hall****Experiment Description:**Due to the sunny weather, Dr. Shaw has granted permission to perform the next lab experiment outdoors. Instead of dropping weights indoors, the summer experiment involves dropping water balloons, with the condition that all balloon fragments are cleaned up afterward.**Experiment Instructions:**For one trial, a water balloon with a mass of 283 grams (0.283 kg) is released from the top of Stanley Hall, landing on the sidewalk below. To calculate the height of Stanley Hall, consider that the balloon travels at a speed of 11.22 m/s when it strikes a lab partner's head, which is positioned 1.44 meters above the sidewalk.**Problem Statement:**Calculate the height of Stanley Hall using the given data:- **Mass of the water balloon**: 283 g (0.283 kg)- **Speed of the balloon on impact**: 11.22 m/s- **Height of the lab partner's head above the sidewalk**: 1.44 meters**Solution:**To find the height of Stanley Hall, we can utilize the principles of kinematic equations for free-falling objects under the influence of gravity.**Kinematic Equation**:\[ v^2 = u^2 + 2gh \]Where:- \( v \) = final velocity (11.22 m/s)- \( u \) = initial velocity (0 m/s, since the balloon is dropped)- \( g \) = acceleration due to gravity (9.81 m/s²)- \( h \) = height from which the balloon is droppedRearrange the kinematic equation to solve for \( h \):\[ h = \frac{v^2 - u^2}{2g} \]Substituting the values:\[ h = \frac{(11.22 \, \text{m/s})^2 - (0 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]\[ h = \frac{126.03 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]\[ h \approx 6.43 \, \text{meters} \]Since this calculation gives us the distance from the lab

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For one data trial, you release a water balloon of mass 283 g from the top of a Stanley Hall onto the sidewalk below. If the balloon is traveling at 11.22 m/s when it strikes your lab partner's head (1.44 m above the sidewalk) how tall is Stanley Hall?

For one data trial, you release a water balloon of mass 283 g from the top of a Stanley Hall onto the sidewalk below. If the balloon is traveling at 11.22 m/s when it strikes your lab partner's head (1.44 m above the sidewalk) how tall is Stanley Hall?

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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Question

**Physics Lab Experiment: Calculating the Height of Stanley Hall**

**Experiment Description:**

Due to the sunny weather, Dr. Shaw has granted permission to perform the next lab experiment outdoors. Instead of dropping weights indoors, the summer experiment involves dropping water balloons, with the condition that all balloon fragments are cleaned up afterward.

**Experiment Instructions:**

For one trial, a water balloon with a mass of 283 grams (0.283 kg) is released from the top of Stanley Hall, landing on the sidewalk below. To calculate the height of Stanley Hall, consider that the balloon travels at a speed of 11.22 m/s when it strikes a lab partner's head, which is positioned 1.44 meters above the sidewalk.

**Problem Statement:**

Calculate the height of Stanley Hall using the given data:
- **Mass of the water balloon**: 283 g (0.283 kg)
- **Speed of the balloon on impact**: 11.22 m/s
- **Height of the lab partner's head above the sidewalk**: 1.44 meters

**Solution:**

To find the height of Stanley Hall, we can utilize the principles of kinematic equations for free-falling objects under the influence of gravity.

**Kinematic Equation**:
\[ v^2 = u^2 + 2gh \]

Where:
- \( v \) = final velocity (11.22 m/s)
- \( u \) = initial velocity (0 m/s, since the balloon is dropped)
- \( g \) = acceleration due to gravity (9.81 m/s²)
- \( h \) = height from which the balloon is dropped

Rearrange the kinematic equation to solve for \( h \):
\[ h = \frac{v^2 - u^2}{2g} \]

Substituting the values:
\[ h = \frac{(11.22 \, \text{m/s})^2 - (0 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]

\[ h = \frac{126.03 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]

\[ h \approx 6.43 \, \text{meters} \]

Since this calculation gives us the distance from the lab

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