A helicopter is ascending vertically with a speed of `8.0 "ms"^(-1)`. At a height of 12 m above the earth, a package is dropped from a window. How much time does it take for the package to reach the ground ?

Options

A. 1.23 s
B. 3.23s
C. 5.83s
D. 2.53s

Answer

To solve the problem of how much time it takes for a package to reach the ground after being dropped from a helicopter ascending vertically, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The helicopter is ascending with a speed of \(8.0 \, \text{m/s}\) at a height of \(12 \, \text{m}\). When the package is dropped, it has the same initial velocity as the helicopter, which is \(8.0 \, \text{m/s}\) upwards.

### Step 2: Set up the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \(s\) = displacement (which will be \(-12 \, \text{m}\) since it is falling down)
- \(u\) = initial velocity (\(8.0 \, \text{m/s}\) upwards, but we will take it as positive)
- \(a\) = acceleration due to gravity (\(-10 \, \text{m/s}^2\), negative because it acts downwards)
- \(t\) = time in seconds

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ -12 = 8t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ -12 = 8t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 8t - 12 = 0 \]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 5\), \(b = -8\), and \(c = -12\).

Calculating the discriminant:
\[ b^2 - 4ac = (-8)^2 - 4 \cdot 5 \cdot (-12) = 64 + 240 = 304 \]

Now substituting back into the quadratic formula:
\[ t = \frac{-(-8) \pm \sqrt{304}}{2 \cdot 5} = \frac{8 \pm \sqrt{304}}{10} \]

### Step 5: Calculate the value of \(t\)
Calculating \(\sqrt{304}\):
\[ \sqrt{304} \approx 17.43 \]
Thus:
\[ t = \frac{8 \pm 17.43}{10} \]

Calculating the two possible values of \(t\):
1. \(t = \frac{8 + 17.43}{10} = \frac{25.43}{10} \approx 2.543 \, \text{s}\)
2. \(t = \frac{8 - 17.43}{10} = \frac{-9.43}{10} \approx -0.943 \, \text{s}\) (not physically meaningful)

### Final Answer
The time it takes for the package to reach the ground is approximately \(2.54 \, \text{s}\).

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