A balloon is going upwards with velocity `12 m//sec` it releases a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground `(g=10 m//sec^(2))`

Options

A. 5 sec
B. 6 sec
C. 7 sec
D. 8 sec

Explanation

To solve the problem of how much time the packet will take to reach the ground after being released from the balloon, we can follow these steps:

### Step 1: Identify the known values
- Initial velocity of the packet (u) = 12 m/s (upwards)
- Height from which the packet is released (s) = -65 m (downward displacement)
- Acceleration due to gravity (g) = -10 m/s² (downward)

### Step 2: Set up the equation of motion
We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = displacement
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ -65 = 12t + \frac{1}{2}(-10)t^2 \]
This simplifies to:
\[ -65 = 12t - 5t^2 \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 5t^2 - 12t - 65 = 0 \]

### Step 5: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 5 \)
- \( b = -12 \)
- \( c = -65 \)

Calculating the discriminant:
\[ b^2 - 4ac = (-12)^2 - 4(5)(-65) = 144 + 1300 = 1444 \]

Now, substituting into the quadratic formula:
\[ t = \frac{12 \pm \sqrt{1444}}{10} \]
\[ \sqrt{1444} = 38 \]
Thus:
\[ t = \frac{12 \pm 38}{10} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{50}{10} = 5 \) seconds (valid)
2. \( t = \frac{-26}{10} = -2.6 \) seconds (not valid since time cannot be negative)

### Conclusion
The packet will take **5 seconds** to reach the ground.

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