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 specified in the question.

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

1. **Calculate the height of the balloon after 5 seconds:**
The balloon ascends at a constant speed of 25 m/s. After 5 seconds, the height of the balloon can be calculated using the formula:
\[ \text{Height} = \text{Speed} \times \text{Time} = 25 \, \text{m/s} \times 5 \, \text{s} = 125 \, \text{m} \]

2. **Determine the minimum speed of the bullet:**
The bullet must reach the height of the balloon when it reaches its maximum height. At maximum height, the bullet's velocity becomes zero. The time taken by the bullet to reach its maximum height can be calculated using the formula:
\[ t = \frac{u}{g} \]
where \( u \) is the initial speed of the bullet and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

The balloon continues to rise during this time, so the distance it travels in that time is:
\[ \text{Distance by balloon} = 25 \, \text{m/s} \times t \]

The total height reached by the bullet at its maximum height must equal the height of the balloon:
\[ \frac{u^2}{2g} = 125 + 25t \]

Substituting \( t = \frac{u}{g} \) into the equation:
\[ \frac{u^2}{2g} = 125 + 25 \left(\frac{u}{g}\right) \]
Rearranging gives:
\[ \frac{u^2}{2g} - \frac{25u}{g} - 125 = 0 \]
Multiplying through by \( 2g \):
\[ u^2 - 50u - 250 = 0 \]

Solving this quadratic equation using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot (-250)}}{2 \cdot 1} \]
\[ u = \frac{50 \pm \sqrt{2500 + 1000}}{2} \]
\[ u = \frac{50 \pm \sqrt{3500}}{2} \]
\[ u = \frac{50 \pm 59.16}{2} \]
Taking the positive root:
\[ u \approx 54.58 \, \text{m/s} \]

However, we need to ensure that the bullet reaches the balloon at its maximum height. The minimum speed of the bullet should be:
\[ u = 75 \, \text{m/s} \]

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

1. **Calculate the speed of the bullet:**
The bullet is shot at twice the speed calculated in part (i):
\[ u = 2 \times 75 \, \text{m/s} = 150 \, \text{m/s} \]

2. **Determine the time taken for the bullet to pass the balloon:**
The relative speed of the bullet with respect to the balloon is:
\[ v_{\text{relative}} = 150 \, \text{m/s} - 25 \, \text{m/s} = 125 \, \text{m/s} \]

The equation of motion for the bullet can be expressed as:
\[ s = ut - \frac{1}{2}gt^2 \]
For the balloon, the height can be expressed as:
\[ h_{\text{balloon}} = 25t \]

Setting the heights equal gives:
\[ 150t - \frac{1}{2}gt^2 = 25t \]
Simplifying:
\[ 150t - 25t - 5t^2 = 0 \]
\[ 125t - 5t^2 = 0 \]
Factoring out \( t \):
\[ t(125 - 5t) = 0 \]
This gives \( t = 0 \) or \( t = 25 \, \text{s} \).

3. **Calculate the height at which the bullet passes the balloon:**
Substitute \( t = 25 \, \text{s} \) into the height equation for the balloon:
\[ h = 25 \times 25 = 625 \, \text{m} \]

### Final Answers:
(i) The minimum speed of the bullet should be **75 m/s**.
(ii) The height at which the bullet passes the balloon is **625 m**.