A balloon starts ascending from the ground at a constant speed of `25 m//s`. After 5 s, a bullet is shot vertically upwards from the ground.
(i) What should be the minimun speed of the bullet so that it may reach the balloon?
(ii) The bullet is shot at twise the speed calculated in (i). Find the height at which it passes the balloon.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (i): Minimum Speed of the Bullet

1. **Determine the height of the balloon after 5 seconds:**
The balloon ascends at a constant speed of 25 m/s. After 5 seconds, the height \( h_b \) of the balloon can be calculated as:
\[ h_b = \text{speed} \times \text{time} = 25 \, \text{m/s} \times 5 \, \text{s} = 125 \, \text{m} \]

2. **Set up the equations for the bullet:**
Let the speed of the bullet be \( v \). The bullet is shot 5 seconds after the balloon starts ascending. The time taken by the bullet to reach the height of the balloon is \( t \).

3. **Height of the bullet when it reaches the balloon:**
The height \( h_b \) of the balloon after \( t \) seconds (when the bullet is shot) is:
\[ h_b = 125 \, \text{m} + 25 \, \text{m/s} \times t \]
The bullet's height \( h_{bullet} \) after \( t \) seconds is given by:
\[ h_{bullet} = v \cdot t - \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

4. **Equate the heights:**
For the bullet to reach the balloon, the heights must be equal:
\[ 125 + 25t = vt - \frac{1}{2} g t^2 \]

5. **Substituting \( g = 10 \, \text{m/s}^2 \):**
\[ 125 + 25t = vt - 5t^2 \]

6. **Rearranging the equation:**
\[ 5t^2 + (v - 25)t - 125 = 0 \]

7. **Using the quadratic formula:**
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = v - 25 \), and \( c = -125 \).

8. **Finding the discriminant:**
For the bullet to just reach the balloon, the discriminant must be zero:
\[ (v - 25)^2 - 4 \cdot 5 \cdot (-125) = 0 \]
Simplifying gives:
\[ (v - 25)^2 + 2500 = 0 \]

9. **Solving for \( v \):**
\[ (v - 25)^2 = 2500 \implies v - 25 = \sqrt{2500} \implies v - 25 = 50 \implies v = 75 \, \text{m/s} \]

### Part (ii): Height at which the Bullet Passes the Balloon

1. **Bullet speed is doubled:**
The bullet is now shot at \( 2v = 150 \, \text{m/s} \).

2. **Using the same height equations:**
The height of the bullet after time \( t \) is:
\[ h_{bullet} = 150t - 5t^2 \]
The height of the balloon after \( t \) seconds is still:
\[ h_b = 125 + 25t \]

3. **Setting the heights equal:**
\[ 150t - 5t^2 = 125 + 25t \]

4. **Rearranging the equation:**
\[ -5t^2 + 125t - 125 = 0 \]
Dividing through by -5 gives:
\[ t^2 - 25t + 25 = 0 \]

5. **Using the quadratic formula again:**
\[ t = \frac{25 \pm \sqrt{25^2 - 4 \cdot 1 \cdot 25}}{2 \cdot 1} = \frac{25 \pm \sqrt{625 - 100}}{2} = \frac{25 \pm \sqrt{525}}{2} \]
Simplifying gives:
\[ t = \frac{25 \pm 5\sqrt{21}}{2} \]

6. **Finding the height at which they meet:**
Substitute \( t \) back into the balloon's height equation:
\[ h_b = 125 + 25 \left(\frac{25 \pm 5\sqrt{21}}{2}\right) \]
This will give the height at which the bullet passes the balloon.

### Final Answers:
(i) Minimum speed of the bullet: **75 m/s**
(ii) Height at which it passes the balloon: **Calculate using the above equations.**