Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Ask your questions and receive precise answers from experienced professionals across different disciplines. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
