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Sagot :
Sure, let's work through each part of the question step-by-step.
### Part a: Finding the Value of [tex]\( U \)[/tex]
- Given:
- The acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex].
- The speed of the stone when it hits the ground [tex]\( 40 \, \text{m/s} \)[/tex].
- Approach:
- Using the conservation of energy principle, the potential energy at the highest point is converted into kinetic energy when the stone hits the ground.
- The kinetic energy at the ground is given as [tex]\( 0.5 m \times (\text{speed at the ground})^2 \)[/tex].
- The potential energy at the highest point is [tex]\( mgh \)[/tex], where [tex]\( h \)[/tex] is the height achieved above point A.
- Since the stone hits the ground with the same speed it had initially but in the opposite direction, the initial speed [tex]\( U \)[/tex] at point A should be equal to the speed it has just before hitting the ground.
- Solution:
[tex]\[ U = \text{speed at the ground} = 40 \, \text{m/s} \][/tex]
The value of [tex]\( U \)[/tex] is [tex]\( 40 \, \text{m/s} \)[/tex].
### Part b: Finding the Value of [tex]\( T \)[/tex]
- Given:
- Initial speed [tex]\( U = 40 \, \text{m/s} \)[/tex]
- Acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Approach:
- Using the kinematic equation for vertical motion.
- The time taken to reach the highest point ([tex]\( T \)[/tex]) can be found using:
[tex]\[ 0 = U - gT \][/tex]
- Solving for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{U}{g} \][/tex]
- Solution:
[tex]\[ T = \frac{40}{9.8} \approx 4.081632653061225 \, \text{s} \][/tex]
The value of [tex]\( T \)[/tex] is approximately [tex]\( 4.081632653061225 \, \text{s} \)[/tex].
### Part c: Suggesting One More Refinement
- One refinement to make this model more realistic, apart from air resistance, would be to consider the change in gravitational acceleration with height. However, for small heights compared to the Earth's radius, this effect is negligible. Additionally, one could consider the effect of the Earth's rotation.
### Part d: Effect of Air Resistance
- Explanation:
- In reality, the stone will not move freely under gravity and will be subject to air resistance.
- Air resistance would reduce both the maximum height reached by the stone and the speed at which it hits the ground.
- Therefore, the value of [tex]\( U \)[/tex] would be less than [tex]\( 40 \, \text{m/s} \)[/tex] when air resistance is considered.
The air resistance would result in the stone having a lower initial speed [tex]\( U \)[/tex] than [tex]\( 40 \, \text{m/s} \)[/tex] due to the energy being dissipated as it moves upwards and also as it descends back towards the ground.
### Part a: Finding the Value of [tex]\( U \)[/tex]
- Given:
- The acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex].
- The speed of the stone when it hits the ground [tex]\( 40 \, \text{m/s} \)[/tex].
- Approach:
- Using the conservation of energy principle, the potential energy at the highest point is converted into kinetic energy when the stone hits the ground.
- The kinetic energy at the ground is given as [tex]\( 0.5 m \times (\text{speed at the ground})^2 \)[/tex].
- The potential energy at the highest point is [tex]\( mgh \)[/tex], where [tex]\( h \)[/tex] is the height achieved above point A.
- Since the stone hits the ground with the same speed it had initially but in the opposite direction, the initial speed [tex]\( U \)[/tex] at point A should be equal to the speed it has just before hitting the ground.
- Solution:
[tex]\[ U = \text{speed at the ground} = 40 \, \text{m/s} \][/tex]
The value of [tex]\( U \)[/tex] is [tex]\( 40 \, \text{m/s} \)[/tex].
### Part b: Finding the Value of [tex]\( T \)[/tex]
- Given:
- Initial speed [tex]\( U = 40 \, \text{m/s} \)[/tex]
- Acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Approach:
- Using the kinematic equation for vertical motion.
- The time taken to reach the highest point ([tex]\( T \)[/tex]) can be found using:
[tex]\[ 0 = U - gT \][/tex]
- Solving for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{U}{g} \][/tex]
- Solution:
[tex]\[ T = \frac{40}{9.8} \approx 4.081632653061225 \, \text{s} \][/tex]
The value of [tex]\( T \)[/tex] is approximately [tex]\( 4.081632653061225 \, \text{s} \)[/tex].
### Part c: Suggesting One More Refinement
- One refinement to make this model more realistic, apart from air resistance, would be to consider the change in gravitational acceleration with height. However, for small heights compared to the Earth's radius, this effect is negligible. Additionally, one could consider the effect of the Earth's rotation.
### Part d: Effect of Air Resistance
- Explanation:
- In reality, the stone will not move freely under gravity and will be subject to air resistance.
- Air resistance would reduce both the maximum height reached by the stone and the speed at which it hits the ground.
- Therefore, the value of [tex]\( U \)[/tex] would be less than [tex]\( 40 \, \text{m/s} \)[/tex] when air resistance is considered.
The air resistance would result in the stone having a lower initial speed [tex]\( U \)[/tex] than [tex]\( 40 \, \text{m/s} \)[/tex] due to the energy being dissipated as it moves upwards and also as it descends back towards the ground.
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