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Sagot :
Alright, let's break down the problem and solve it step-by-step.
### Given:
- Mass of the ball ([tex]\( m \)[/tex]) = 0.1 kg (converted from 0.1tg)
- Initial velocity ([tex]\( u \)[/tex]) = 80 m/s
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
### 1. Calculate the Maximum Height
The kinematic equation we will use is:
[tex]\[ v^2 = u^2 - 2gh \][/tex]
Where:
- [tex]\( v \)[/tex] = final velocity (0 m/s at maximum height)
- [tex]\( h \)[/tex] = maximum height
Rearranging for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{u^2}{2g} \][/tex]
### 2. Calculate the Potential Energy at Half the Maximum Height
First, we find the maximum height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{(80 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ h = \frac{6400}{19.6} \][/tex]
[tex]\[ h = 326.53 \, \text{m} \][/tex]
Now, the half of the maximum height [tex]\( h/2 \)[/tex] is:
[tex]\[ \frac{h}{2} = \frac{326.53 \, \text{m}}{2} \][/tex]
[tex]\[ \frac{h}{2} = 163.26 \, \text{m} \][/tex]
The potential energy at half the height is:
[tex]\[ PE_{\text{halfway}} = mgh_{\text{half}} \][/tex]
[tex]\[ PE_{\text{halfway}} = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 163.26 \, \text{m} \][/tex]
[tex]\[ PE_{\text{halfway}} = 159.99 \, \text{J} \][/tex]
Approximately, [tex]\( 160 \, \text{J} \)[/tex].
### 3. Calculate the Potential Energy at Its Maximum Height
The potential energy at maximum height is:
[tex]\[ PE_{\text{max}} = mgh \][/tex]
[tex]\[ PE_{\text{max}} = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 326.53 \, \text{m} \][/tex]
[tex]\[ PE_{\text{max}} = 319.99 \, \text{J} \][/tex]
Approximately, [tex]\( 320 \, \text{J} \)[/tex].
### 4. Calculate the Kinetic Energy as it Leaves the Ground
The kinetic energy as it leaves the ground is given by the formula:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]
Substituting the given values:
[tex]\[ KE = \frac{1}{2} \times 0.1 \, \text{kg} \times (80 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = 0.05 \times 6400 \][/tex]
[tex]\[ KE = 320 \, \text{J} \][/tex]
### Summary
1. Potential Energy halfway up: [tex]\( 160 \, \text{J} \)[/tex]
2. Potential Energy at maximum height: [tex]\( 320 \, \text{J} \)[/tex]
3. Kinetic Energy as it leaves the ground: [tex]\( 320 \, \text{J} \)[/tex]
These are the detailed calculations and the results for each part of the problem.
### Given:
- Mass of the ball ([tex]\( m \)[/tex]) = 0.1 kg (converted from 0.1tg)
- Initial velocity ([tex]\( u \)[/tex]) = 80 m/s
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
### 1. Calculate the Maximum Height
The kinematic equation we will use is:
[tex]\[ v^2 = u^2 - 2gh \][/tex]
Where:
- [tex]\( v \)[/tex] = final velocity (0 m/s at maximum height)
- [tex]\( h \)[/tex] = maximum height
Rearranging for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{u^2}{2g} \][/tex]
### 2. Calculate the Potential Energy at Half the Maximum Height
First, we find the maximum height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{(80 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ h = \frac{6400}{19.6} \][/tex]
[tex]\[ h = 326.53 \, \text{m} \][/tex]
Now, the half of the maximum height [tex]\( h/2 \)[/tex] is:
[tex]\[ \frac{h}{2} = \frac{326.53 \, \text{m}}{2} \][/tex]
[tex]\[ \frac{h}{2} = 163.26 \, \text{m} \][/tex]
The potential energy at half the height is:
[tex]\[ PE_{\text{halfway}} = mgh_{\text{half}} \][/tex]
[tex]\[ PE_{\text{halfway}} = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 163.26 \, \text{m} \][/tex]
[tex]\[ PE_{\text{halfway}} = 159.99 \, \text{J} \][/tex]
Approximately, [tex]\( 160 \, \text{J} \)[/tex].
### 3. Calculate the Potential Energy at Its Maximum Height
The potential energy at maximum height is:
[tex]\[ PE_{\text{max}} = mgh \][/tex]
[tex]\[ PE_{\text{max}} = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 326.53 \, \text{m} \][/tex]
[tex]\[ PE_{\text{max}} = 319.99 \, \text{J} \][/tex]
Approximately, [tex]\( 320 \, \text{J} \)[/tex].
### 4. Calculate the Kinetic Energy as it Leaves the Ground
The kinetic energy as it leaves the ground is given by the formula:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]
Substituting the given values:
[tex]\[ KE = \frac{1}{2} \times 0.1 \, \text{kg} \times (80 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = 0.05 \times 6400 \][/tex]
[tex]\[ KE = 320 \, \text{J} \][/tex]
### Summary
1. Potential Energy halfway up: [tex]\( 160 \, \text{J} \)[/tex]
2. Potential Energy at maximum height: [tex]\( 320 \, \text{J} \)[/tex]
3. Kinetic Energy as it leaves the ground: [tex]\( 320 \, \text{J} \)[/tex]
These are the detailed calculations and the results for each part of the problem.
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