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Sagot :
### Problem:
We have a ball hit vertically upward with an initial velocity of [tex]\(70 \, \text{m/s}\)[/tex]. We need to determine:
(a) the maximum height reached,
(b) the time taken to reach the maximum height,
(c) the total time of flight,
(d) the parameters to consider for a free-falling body.
### Solution:
1. Given Data:
- Initial velocity, [tex]\( u = 70 \, \text{m/s} \)[/tex]
- Acceleration due to gravity, [tex]\( g = -9.81 \, \text{m/s}^2 \)[/tex] (negative because it's acting downward)
2. Equations and Concepts:
- To find the maximum height, we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
[tex]\( v \)[/tex] = final velocity at max height (which is [tex]\( 0 \, \text{m/s} \)[/tex]),
[tex]\( a \)[/tex] = acceleration due to gravity [tex]\( (g) \)[/tex],
[tex]\( s \)[/tex] = maximum height.
- To find the time to reach maximum height, we use:
[tex]\[ v = u + at \][/tex]
Where:
[tex]\( t \)[/tex] = time taken to reach max height.
- The total time of flight is double the time taken to reach the maximum height, as the time to ascend and descend is the same.
### Calculations:
(a) Maximum Height Reached:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Setting [tex]\( v = 0 \, \text{m/s} \)[/tex], we solve for [tex]\( s \)[/tex]:
[tex]\[ 0 = (70)^2 + 2 \times (-9.81) \times s \][/tex]
[tex]\[ 0 = 4900 - 19.62s \][/tex]
[tex]\[ 19.62s = 4900 \][/tex]
[tex]\[ s = \frac{4900}{19.62} \][/tex]
[tex]\[ s \approx 249.75 \, \text{m} \][/tex]
(b) Time to Reach Maximum Height:
[tex]\[ v = u + at \][/tex]
[tex]\[ 0 = 70 + (-9.81)t \][/tex]
[tex]\[ -70 = -9.81t \][/tex]
[tex]\[ t = \frac{70}{9.81} \][/tex]
[tex]\[ t \approx 7.14 \, \text{s} \][/tex]
(c) Total Time of Flight:
The total time is twice the time to reach the maximum height:
[tex]\[ \text{Total time} = 2 \times 7.14 = 14.27 \, \text{s} \][/tex]
(d) Parameters to Consider for Free-Falling Body:
1. Initial velocity (u): The starting speed of the object.
2. Final velocity (v): The speed of the object at a specific point or when it comes to rest.
3. Acceleration due to gravity (g): The constant rate at which objects accelerate downwards due to Earth's gravity, typically [tex]\( -9.81 \, \text{m/s}^2 \)[/tex].
4. Displacement (s): The change in position of the object.
5. Time (t): The duration for which the object is in motion.
### Summary:
- The maximum height reached by the ball is approximately 249.75 meters.
- The time taken to reach the maximum height is approximately 7.14 seconds.
- The total time of flight is approximately 14.27 seconds.
- The parameters to consider for a free-falling body include initial velocity, final velocity, acceleration due to gravity, displacement, and time.
We have a ball hit vertically upward with an initial velocity of [tex]\(70 \, \text{m/s}\)[/tex]. We need to determine:
(a) the maximum height reached,
(b) the time taken to reach the maximum height,
(c) the total time of flight,
(d) the parameters to consider for a free-falling body.
### Solution:
1. Given Data:
- Initial velocity, [tex]\( u = 70 \, \text{m/s} \)[/tex]
- Acceleration due to gravity, [tex]\( g = -9.81 \, \text{m/s}^2 \)[/tex] (negative because it's acting downward)
2. Equations and Concepts:
- To find the maximum height, we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
[tex]\( v \)[/tex] = final velocity at max height (which is [tex]\( 0 \, \text{m/s} \)[/tex]),
[tex]\( a \)[/tex] = acceleration due to gravity [tex]\( (g) \)[/tex],
[tex]\( s \)[/tex] = maximum height.
- To find the time to reach maximum height, we use:
[tex]\[ v = u + at \][/tex]
Where:
[tex]\( t \)[/tex] = time taken to reach max height.
- The total time of flight is double the time taken to reach the maximum height, as the time to ascend and descend is the same.
### Calculations:
(a) Maximum Height Reached:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Setting [tex]\( v = 0 \, \text{m/s} \)[/tex], we solve for [tex]\( s \)[/tex]:
[tex]\[ 0 = (70)^2 + 2 \times (-9.81) \times s \][/tex]
[tex]\[ 0 = 4900 - 19.62s \][/tex]
[tex]\[ 19.62s = 4900 \][/tex]
[tex]\[ s = \frac{4900}{19.62} \][/tex]
[tex]\[ s \approx 249.75 \, \text{m} \][/tex]
(b) Time to Reach Maximum Height:
[tex]\[ v = u + at \][/tex]
[tex]\[ 0 = 70 + (-9.81)t \][/tex]
[tex]\[ -70 = -9.81t \][/tex]
[tex]\[ t = \frac{70}{9.81} \][/tex]
[tex]\[ t \approx 7.14 \, \text{s} \][/tex]
(c) Total Time of Flight:
The total time is twice the time to reach the maximum height:
[tex]\[ \text{Total time} = 2 \times 7.14 = 14.27 \, \text{s} \][/tex]
(d) Parameters to Consider for Free-Falling Body:
1. Initial velocity (u): The starting speed of the object.
2. Final velocity (v): The speed of the object at a specific point or when it comes to rest.
3. Acceleration due to gravity (g): The constant rate at which objects accelerate downwards due to Earth's gravity, typically [tex]\( -9.81 \, \text{m/s}^2 \)[/tex].
4. Displacement (s): The change in position of the object.
5. Time (t): The duration for which the object is in motion.
### Summary:
- The maximum height reached by the ball is approximately 249.75 meters.
- The time taken to reach the maximum height is approximately 7.14 seconds.
- The total time of flight is approximately 14.27 seconds.
- The parameters to consider for a free-falling body include initial velocity, final velocity, acceleration due to gravity, displacement, and time.
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