A ballon is ascending at the rate `v = 12 km//h` and is being carried horizontally by the wind at `v_(w) = 20 km//h`. If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the ballon with the same velocity as the balloon. Also, find the speed with which the bag strikes the ground?

Explanation

To solve the problem step by step, we will first convert the velocities from km/h to m/s, then use the equations of motion to find the time taken for the ballast bag to hit the ground and finally calculate the speed at which it strikes the ground.

### Step-by-Step Solution:

1. **Convert Velocities from km/h to m/s:**
- The ascending velocity of the balloon, \( v = 12 \, \text{km/h} \)
- The wind velocity, \( v_w = 20 \, \text{km/h} \)

To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \):
\[ v = 12 \times \frac{5}{18} = 3.33 \, \text{m/s} \]
\[ v_w = 20 \times \frac{5}{18} = 5.55 \, \text{m/s} \]

2. **Identify the Initial Conditions:**
- The initial vertical velocity of the ballast bag when dropped, \( u_y = 3.33 \, \text{m/s} \) (upward).
- The height from which the bag is dropped, \( h = 50 \, \text{m} \).
- The acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (acting downward).

3. **Use the Equation of Motion to Find Time:**
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = -50 \, \text{m} \) (downward), \( u = 3.33 \, \text{m/s} \), and \( a = -g = -10 \, \text{m/s}^2 \).
Plugging in the values:
\[ -50 = 3.33t - \frac{1}{2} \cdot 10 t^2 \]
Rearranging gives:
\[ 5t^2 - 3.33t - 50 = 0 \]

4. **Solve the Quadratic Equation:**
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = -3.33 \), \( c = -50 \).
\[ t = \frac{-(-3.33) \pm \sqrt{(-3.33)^2 - 4 \cdot 5 \cdot (-50)}}{2 \cdot 5} \]
\[ t = \frac{3.33 \pm \sqrt{11.0889 + 1000}}{10} \]
\[ t = \frac{3.33 \pm \sqrt{1011.0889}}{10} \]
\[ t = \frac{3.33 \pm 31.78}{10} \]
Taking the positive root:
\[ t = \frac{35.11}{10} = 3.51 \, \text{s} \]

5. **Calculate the Final Velocity:**
The final vertical velocity \( v_y \) when the bag strikes the ground can be calculated using:
\[ v_y = u_y - gt \]
\[ v_y = 3.33 - 10 \times 3.51 \]
\[ v_y = 3.33 - 35.1 = -31.77 \, \text{m/s} \]

6. **Calculate the Resultant Velocity:**
The resultant velocity \( V \) when it strikes the ground can be calculated using:
\[ V = \sqrt{v_x^2 + v_y^2} \]
Where \( v_x = 5.55 \, \text{m/s} \) (horizontal component):
\[ V = \sqrt{(5.55)^2 + (-31.77)^2} \]
\[ V = \sqrt{30.80 + 1009.76} = \sqrt{1040.56} \approx 32.25 \, \text{m/s} \]

### Final Answers:
- **Time taken to strike the ground:** \( t \approx 3.51 \, \text{s} \)
- **Speed with which the bag strikes the ground:** \( V \approx 32.25 \, \text{m/s} \)