A balloon carring a stone is moving vartically upward with velocity `12 m//s`. When the balloon is at height `64 m`, the stone is dropped. After how much time and with what velocity will it strike the ground?

Options

A. `4 sec.`
B. `5 sec.`
C. `6 sec.`
D. None of these

Explanation

To solve the problem, we need to determine the time it takes for the stone to hit the ground after being dropped from a height of 64 meters while moving upward with an initial velocity of 12 m/s. We will also find the velocity of the stone just before it strikes the ground.

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial height (S) = 64 m (upward direction taken as negative)
- Initial velocity (U) = -12 m/s (since the stone is moving upward)
- Acceleration due to gravity (A) = 10 m/s² (downward direction taken as positive)

2. **Set up the equation of motion:**
We will use the equation of motion:
\[ S = Ut + \frac{1}{2} A t^2 \]
Substituting the known values:
\[ 64 = -12t + \frac{1}{2} \times 10 \times t^2 \]
This simplifies to:
\[ 64 = -12t + 5t^2 \]

3. **Rearranging the equation:**
Rearranging gives us a standard quadratic equation:
\[ 5t^2 - 12t - 64 = 0 \]

4. **Applying the quadratic formula:**
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -12 \), and \( c = -64 \). Plugging in these values:
\[ t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot (-64)}}{2 \cdot 5} \]

5. **Calculating the discriminant:**
First, calculate \( b^2 - 4ac \):
\[ (-12)^2 = 144 \]
\[ -4 \cdot 5 \cdot (-64) = 1280 \]
Therefore:
\[ b^2 - 4ac = 144 + 1280 = 1424 \]

6. **Finding the square root:**
Now, calculate the square root:
\[ \sqrt{1424} \approx 37.7 \]

7. **Finding the time (t):**
Substitute back into the quadratic formula:
\[ t = \frac{12 \pm 37.7}{10} \]
This gives two possible values for t:
\[ t_1 = \frac{49.7}{10} \approx 4.97 \text{ seconds} \]
\[ t_2 = \frac{-25.7}{10} \text{ (not valid as time cannot be negative)} \]
So, we take \( t \approx 5 \text{ seconds} \).

8. **Finding the final velocity (V):**
Now, we will use the formula:
\[ V = U + At \]
Substituting the values:
\[ V = -12 + 10 \cdot 5 \]
\[ V = -12 + 50 = 38 \text{ m/s} \]

### Final Answers:
- Time taken to strike the ground: **5 seconds**
- Velocity just before striking the ground: **38 m/s**