Answer :
To find the maximum height reached by the ball, we can use one of the kinematic equations of motion. The main variables involved are the initial velocity, the acceleration due to gravity, and the final velocity at the maximum height.
1. Initial Information:
- Mass, [tex]\( m \)[/tex] = 0.25 kg (though this is not needed for our calculations)
- Initial velocity, [tex]\( u \)[/tex] = 12 m/s
- Final velocity at maximum height, [tex]\( v \)[/tex] = 0 m/s (since the ball momentarily stops at the highest point)
- Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s² (downwards, so it will be negative in our upward motion equation)
2. Choosing the Right Kinematic Equation:
The relevant kinematic equation that relates these variables is:
[tex]\[ v^2 = u^2 + 2as \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( s \)[/tex] is the displacement or height
3. Rearrange the Equation to Solve for Height [tex]\( s \)[/tex]:
Since we're finding the maximum height, and knowing that [tex]\( v = 0 \)[/tex] at the highest point:
[tex]\[ 0 = u^2 + 2as \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ 0 = u^2 + 2(-g)s \][/tex]
[tex]\[ u^2 = 2gs \][/tex]
[tex]\[ s = \frac{-u^2}{2(-g)} \][/tex]
[tex]\[ s = \frac{u^2}{2g} \][/tex]
4. Substitute the Known Values:
[tex]\[ s = \frac{(12 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ s = \frac{144 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \][/tex]
5. Calculate the Height:
[tex]\[ s \approx 7.339449541284403 \, \text{m} \][/tex]
Therefore, the ball reaches a maximum height of approximately [tex]\( 7.35 \, \text{m} \)[/tex] above its original position.
1. Initial Information:
- Mass, [tex]\( m \)[/tex] = 0.25 kg (though this is not needed for our calculations)
- Initial velocity, [tex]\( u \)[/tex] = 12 m/s
- Final velocity at maximum height, [tex]\( v \)[/tex] = 0 m/s (since the ball momentarily stops at the highest point)
- Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s² (downwards, so it will be negative in our upward motion equation)
2. Choosing the Right Kinematic Equation:
The relevant kinematic equation that relates these variables is:
[tex]\[ v^2 = u^2 + 2as \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( s \)[/tex] is the displacement or height
3. Rearrange the Equation to Solve for Height [tex]\( s \)[/tex]:
Since we're finding the maximum height, and knowing that [tex]\( v = 0 \)[/tex] at the highest point:
[tex]\[ 0 = u^2 + 2as \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ 0 = u^2 + 2(-g)s \][/tex]
[tex]\[ u^2 = 2gs \][/tex]
[tex]\[ s = \frac{-u^2}{2(-g)} \][/tex]
[tex]\[ s = \frac{u^2}{2g} \][/tex]
4. Substitute the Known Values:
[tex]\[ s = \frac{(12 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ s = \frac{144 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \][/tex]
5. Calculate the Height:
[tex]\[ s \approx 7.339449541284403 \, \text{m} \][/tex]
Therefore, the ball reaches a maximum height of approximately [tex]\( 7.35 \, \text{m} \)[/tex] above its original position.
