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

Answer

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

### Step 1: Understand the Initial Conditions
The balloon is moving upwards with a velocity of 12 m/s when it releases the packet at a height of 65 m. The initial velocity of the packet (when released) is also 12 m/s upwards.

### Step 2: Set Up the Coordinate System
We will consider the upward direction as positive. Therefore:
- Initial velocity (u) = +12 m/s (upwards)
- Displacement (s) = -65 m (downwards, as it falls to the ground)
- Acceleration (a) = -10 m/s² (due to gravity, acting downwards)

### Step 3: Use the Equation of Motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the known values:
\[ -65 = 12t + \frac{1}{2}(-10)t^2 \]

### Step 4: Simplify the Equation
This simplifies to:
\[ -65 = 12t - 5t^2 \]

Rearranging gives us:
\[ 5t^2 - 12t - 65 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we have a quadratic equation in the standard form \( at^2 + bt + c = 0 \):
- \( a = 5 \)
- \( b = -12 \)
- \( c = -65 \)

Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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

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

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

### Step 6: Conclusion
The valid solution is:
\[ t = 5 \text{ seconds} \]

Thus, the packet will take **5 seconds** to reach the ground.

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