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Sagot :
Let's solve the problem step-by-step:
### Given:
- Initial velocity ([tex]\(u\)[/tex]) = 10.0 m/s (upward)
- Height of the building ([tex]\(h_\text{building}\)[/tex]) = 50.0 m
- Acceleration due to gravity ([tex]\(g\)[/tex]) = 9.81 m/s²
### (a) Velocity with which the ball will strike the ground:
To find the final velocity ([tex]\(v\)[/tex]) when the ball hits the ground, we can use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
- [tex]\(u\)[/tex] is the initial velocity
- [tex]\(a\)[/tex] is the acceleration (gravity, [tex]\(g\)[/tex])
- [tex]\(s\)[/tex] is the displacement (total vertical distance traveled by the ball)
First, let's find the total vertical distance:
#### Step 1: Time to reach the maximum height:
The time it takes to reach the maximum height ([tex]\(t_\text{up}\)[/tex]) can be calculated using:
[tex]\[ t_\text{up} = \frac{u}{g} \][/tex]
#### Step 2: Maximum height above the building:
The maximum height ([tex]\(h_\text{max}\)[/tex]) above the building is given by:
[tex]\[ h_\text{max} = u \cdot t_\text{up} - \frac{1}{2} g \cdot t_\text{up}^2 \][/tex]
#### Step 3: Total height from the ground:
The total height ([tex]\(h_\text{total}\)[/tex]) from the ground is:
[tex]\[ h_\text{total} = h_\text{building} + h_\text{max} \][/tex]
#### Step 4: Final velocity using the kinematic equation:
Now, we can calculate the final velocity when the ball hits the ground using:
[tex]\[ v = \sqrt{u^2 + 2gh_\text{total}} \][/tex]
Given our conditions, the final velocity when the ball strikes the ground is:
[tex]\[ v \approx 32.88 \text{ m/s} \][/tex]
### (b) Time to strike the ground:
To find the total time ([tex]\(t_\text{total}\)[/tex]) it takes for the ball to strike the ground, we consider two intervals:
1. Time to reach the maximum height ([tex]\(t_\text{up}\)[/tex]).
2. Time to fall from the maximum height to the ground.
#### Step 1: Calculate [tex]\( t_\text{up} \)[/tex]:
[tex]\[ t_\text{up} = \frac{u}{g} \][/tex]
#### Step 2: Time to fall from maximum height ([tex]\(t_\text{down}\)[/tex]):
We can use the kinematic equation to find [tex]\( t_\text{down} \)[/tex]:
[tex]\[ t_\text{down} = \sqrt{\frac{2h_\text{total}}{g}} \][/tex]
#### Step 3: Total time ([tex]\( t_\text{total} \)[/tex]):
[tex]\[ t_\text{total} = t_\text{up} + t_\text{down} \][/tex]
Given our conditions, the total time for the ball to hit the ground is:
[tex]\[ t_\text{total} \approx 4.37 \text{ seconds} \][/tex]
### Summary:
(a) The velocity with which the ball will strike the ground is approximately:
[tex]\[ 32.88 \text{ m/s} \][/tex]
(b) The total time it takes for the ball to strike the ground is approximately:
[tex]\[ 4.37 \text{ seconds} \][/tex]
### Given:
- Initial velocity ([tex]\(u\)[/tex]) = 10.0 m/s (upward)
- Height of the building ([tex]\(h_\text{building}\)[/tex]) = 50.0 m
- Acceleration due to gravity ([tex]\(g\)[/tex]) = 9.81 m/s²
### (a) Velocity with which the ball will strike the ground:
To find the final velocity ([tex]\(v\)[/tex]) when the ball hits the ground, we can use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
- [tex]\(u\)[/tex] is the initial velocity
- [tex]\(a\)[/tex] is the acceleration (gravity, [tex]\(g\)[/tex])
- [tex]\(s\)[/tex] is the displacement (total vertical distance traveled by the ball)
First, let's find the total vertical distance:
#### Step 1: Time to reach the maximum height:
The time it takes to reach the maximum height ([tex]\(t_\text{up}\)[/tex]) can be calculated using:
[tex]\[ t_\text{up} = \frac{u}{g} \][/tex]
#### Step 2: Maximum height above the building:
The maximum height ([tex]\(h_\text{max}\)[/tex]) above the building is given by:
[tex]\[ h_\text{max} = u \cdot t_\text{up} - \frac{1}{2} g \cdot t_\text{up}^2 \][/tex]
#### Step 3: Total height from the ground:
The total height ([tex]\(h_\text{total}\)[/tex]) from the ground is:
[tex]\[ h_\text{total} = h_\text{building} + h_\text{max} \][/tex]
#### Step 4: Final velocity using the kinematic equation:
Now, we can calculate the final velocity when the ball hits the ground using:
[tex]\[ v = \sqrt{u^2 + 2gh_\text{total}} \][/tex]
Given our conditions, the final velocity when the ball strikes the ground is:
[tex]\[ v \approx 32.88 \text{ m/s} \][/tex]
### (b) Time to strike the ground:
To find the total time ([tex]\(t_\text{total}\)[/tex]) it takes for the ball to strike the ground, we consider two intervals:
1. Time to reach the maximum height ([tex]\(t_\text{up}\)[/tex]).
2. Time to fall from the maximum height to the ground.
#### Step 1: Calculate [tex]\( t_\text{up} \)[/tex]:
[tex]\[ t_\text{up} = \frac{u}{g} \][/tex]
#### Step 2: Time to fall from maximum height ([tex]\(t_\text{down}\)[/tex]):
We can use the kinematic equation to find [tex]\( t_\text{down} \)[/tex]:
[tex]\[ t_\text{down} = \sqrt{\frac{2h_\text{total}}{g}} \][/tex]
#### Step 3: Total time ([tex]\( t_\text{total} \)[/tex]):
[tex]\[ t_\text{total} = t_\text{up} + t_\text{down} \][/tex]
Given our conditions, the total time for the ball to hit the ground is:
[tex]\[ t_\text{total} \approx 4.37 \text{ seconds} \][/tex]
### Summary:
(a) The velocity with which the ball will strike the ground is approximately:
[tex]\[ 32.88 \text{ m/s} \][/tex]
(b) The total time it takes for the ball to strike the ground is approximately:
[tex]\[ 4.37 \text{ seconds} \][/tex]
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