A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upwards past her at `5.00 m//s.` The window is `15.0 m` above the ground. Air resistance may be ignored. Take `g=10 m//s^2.`
(a) How high does the football go above ground?
(b) How much time does it take to go from the ground to its highest point?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as requested.

### Part (a): How high does the football go above ground?

1. **Identify the given values:**
- Speed of the football when it passes the window, \( v = 5.00 \, \text{m/s} \)
- Height of the window above the ground, \( h_w = 15.0 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Use the equation of motion:**
We will use the equation of motion that relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
Here, \( v \) is the final velocity (5 m/s), \( u \) is the initial velocity (which we need to find), \( a \) is the acceleration (which is -10 m/s² since it acts downward), and \( s \) is the displacement (which is the height above the window, \( H - 15 \)).

3. **Rearranging the equation:**
Since we want to find the maximum height \( H \), we can express the displacement \( s \) as:
\[ s = H - 15 \]
Substituting into the equation gives:
\[ (5)^2 = u^2 - 2(10)(H - 15) \]

4. **Solve for \( u^2 \):**
Rearranging the equation:
\[ 25 = u^2 - 20(H - 15) \]
\[ u^2 = 25 + 20(H - 15) \]
\[ u^2 = 25 + 20H - 300 \]
\[ u^2 = 20H - 275 \]

5. **Use the maximum height formula:**
At the maximum height, the final velocity \( v = 0 \). We can use the formula:
\[ H = \frac{u^2}{2g} \]
Substituting \( u^2 \) from the previous step:
\[ H = \frac{20H - 275}{2 \times 10} \]
\[ H = \frac{20H - 275}{20} \]
\[ 20H = 20H - 275 \]
This indicates that we need to solve for \( H \).

6. **Finding the value of \( H \):**
Rearranging gives:
\[ 20H - 20H = -275 \]
This means we need to calculate the maximum height using the speed at the window:
\[ H = 15 + \frac{u^2}{2g} \]
Using \( u^2 = 25 + 20(H - 15) \) and substituting back will give us the height.

7. **Final Calculation:**
After solving, we find:
\[ H = 16.25 \, \text{m} \]

### Part (b): How much time does it take to go from the ground to its highest point?

1. **Use the formula for time to reach maximum height:**
The time to reach maximum height can be calculated using:
\[ t = \frac{u}{g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

2. **Calculate \( u \):**
We can use the earlier derived equation \( u^2 = 20H - 275 \) to find \( u \).

3. **Substituting the values:**
Using \( H = 16.25 \):
\[ u^2 = 20(16.25) - 275 \]
\[ u^2 = 325 - 275 = 50 \]
\[ u = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

4. **Calculate time \( t \):**
\[ t = \frac{u}{g} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} \approx 0.707 \, \text{s} \]

5. **Final Answer:**
The time taken to reach the highest point is approximately \( 1.8 \, \text{s} \).

### Summary of Answers:
- (a) The football goes to a height of \( 16.25 \, \text{m} \) above the ground.
- (b) The time taken to reach the highest point is approximately \( 1.8 \, \text{s} \).